In isosceles ABC, AB=AC. if B=55,calculate A​

Answer: 1.45

Step-by-step explanation:

Answer:

A) - 54.48

Step-by-step explanation:

Just plug it into the calculator like this;  (-68.1) x 0.8, which equals -54.48.

the answer is negative because you multiplied a positive # and a negative #.

Answer:

use the linear equation of a line

equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.

now u do the rest

Answer:

Step-by-step explanation:

3.8-2.923

Change to 3.800 to match the number of digits in the second number. Work from right to left. 3 minus 0 is 7. work our way over until the end. You should get 0.877 for an answer.

Answer:

That old mans change in height is -1 1/4 over 5 years.

Step-by-step explanation:

Answer:

-5/4 cm would be the answer.

Step-by-step explanation

This is the answer because -1/4 cm multiplied by 5 years would equal -5/4 cm.

Answer:

uh is  it 2 and o.6 i think

Step-by-step explanation:

Answer:First suppose that the roots of the equation

x2−bx+c=0(1)

are real and positive. From the quadratic formula, we see that the roots of (1) are of the form

b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2.

For the root or roots to be real, we require that b2−4c≥0, that is, b2≥4c. In order for them to be positive, we require that

b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0.

This immediately tells us that b>0, but we can go further. We can rearrange this to get

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

which (assuming that b>0) is true if and only if

b2>b2−4c,

since both sides of the inequality are positive so we may square. But then

4c>0.

That is, if the roots are real and positive then b>0 and b2≥4c>0.

Now suppose that b>0 and b2≥4c>0.

Then the roots of (1) are real since b2−4c≥0, and b>0 guarantees that the root b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ is positive.

So it remains to show that b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0. We have that

4c>0,

so that

b2>b2−4c,

then square rooting shows that

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

so the roots of (1) are real and positive, as required.

Approach 2

the curve y equals x squared minus b x plus c showing two positive roots for y equals zero

This is intended to be a proof without words! We have from the diagram that:

If c, b and b2−4c are all positive, there are two real positive roots for x2−bx+c=0 (if b2=4c, we have two real positive equal roots).

If there are two real positive roots for x2−bx+c=0, then c and b are positive and b2−4c is non-negative.

(Why is the distance between the roots b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√?)

Approach 3

When solving problems about the roots of polynomials, it is often useful to find expressions those roots must satisfy and see if this tells us anything new. If α and β denote the roots of the equation, then

x2−bx+c=(x−α)(x−β)=x2−(α+β)x+αβ

and so α+β=b and αβ=c.

We also know that the roots of a quadratic equation are real if and only if the discriminant is non-negative, that is, if and only if b2−4c≥0.

Using these facts, if α and β are both real and positive, then b=α+β>0, c=αβ>0 and b2≥4c, as above.

Conversely, if b>0 and b2≥4c>0, then we know the discriminant is positive and hence both roots are real. We also have that

αβ>0(2)

and

α+β>0.(3)

As α and β are both real, by (2), we know that α and β are either both positive or both negative. However, if α and β were both negative, then (3) could not possibly hold. Hence α and β are both positive, as required.

We now sketch on a graph the region where b>0, c>0 and b2≥4c:

The curve b squared = 4 c as a quadratic with the c-axis vertical and the b-axis horizontal. The region below it is shaded.

The region of the b-c plane for which b>0, c>0 and b2≥4c

Sketch the region of the b-c plane in which the roots of the equation are real and less than 1 in magnitude.

We know that in order for the roots to be real we need b2≥4c as in the first part. We now need to find the region where

−1<b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.(4)

We have b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2≤b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2 so we only need to consider the values for which both

−1<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2andb+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.

Firstly, we will consider the values for which

b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2.

Rearranging gives

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2−b.

So b<2, as the square root is non-negative, and we can square both sides to get

b2−4c<4−4b+b2,

which we may rearrange to find c>b−1.

We will now consider the values for which

−2<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√.

Similarly, we can rearrange to get

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<b+2.

So b>−2, and we can, as before, square to get

b2−4c<b2+4b+4,

and hence c>−b−1.

To sketch the graph, we start by considering the boundary curves b2=4c, c=b−1 and c=−b−1, and the points at which they intersect. We can see that the two lines only intersect when b=0 and c=−1, and the lines intersect the curve when

b2=4b−4andb2=−4b−4

which rearrange to

(b−2)2=0and(b+2)2=0.

This tells us that each line intersects with the curve in only one place and so these lines must be tangent.

Is there a way we could have deduced this directly from (4)?

What do the lines being tangent signify in terms of our equation x2−bx+c=0?

Is this a representation of a well-known property of these equations?

Sketching the graph then yields the following picture:

The graph with the previous curve and the lines 4 c = 4 b minus 4 and 4 c = minus 4 b - 4. Each line touches the curve once and the two lines intersect at (0, minus 1). The region between the three lines/curves is shaded.

The shaded region is where b2≥4c, c>b−1, c>−b−1 and −2<b<2

We might notice that when we derived the inequalities c>b−1 and c>−b−1, a lot of the work we did was very similar.

We might also notice that the graph above is symmetric about the c axis. Why is this?

Does the graph give us any ideas about how we might deduce one inequality from the other?

Step-by-step explanation:

Answer: It contains 88 g of zinc.

Step-by-step explanation:

Given: The ratio of zinc to copper is 4 to 13.

Ratio can be written as [tex]\dfrac4{13}[/tex]

A jar of the chemical contains 286 g of copper.

Let the quantity of zinc be x.

By using the direct proportion rule:[tex]\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}[/tex]

[tex]\dfrac{x}{286}=\dfrac{4}{13}\\\\ x=\dfrac{286\times4}{13}\\\\\Rightarrow\ x=88[/tex]

Hence,  It contains 88 g of zinc.

Answer:

[tex]a = p*q\\b = (p*s)+(q*r)\\c = r*s[/tex]

Step-by-step explanation:

The standard form of trinomial is given as:

[tex]ax^2+bx+c[/tex]

And the factored form is:

[tex](px+r)(qx+s)[/tex]

In order to find the values of a,b and c in terms of p,q,r and s we will take the factored form, multiply it and then compare it with the standard form.

So,

[tex](px+r)(qx+s)\\=px(qx+s)+r(qx+s)\\= pqx^2+psx+qrx+rs\\= pqx^2+(ps+qr)x+rs[/tex]

Now comparing it with the standard form of trinomial

We will compare the co-efficients of x^2, x and the constant

By comparing, we get

[tex]a = pq = p*q\\b = ps+qr = (p*s)+(q*r)\\c = rs = r*s[/tex]

Hence,

[tex]a = p*q\\b = (p*s)+(q*r)\\c = r*s[/tex]

Answer:

w=1.5 or w=13.5

Step-by-step explanation:

this is one answer:

multiply w on both sides to leave u with

4.5=3w

divide both sides by 3

[tex]\frac{4.5}{3}=\frac{3w}{3}[/tex]

1.5=w

_____________________

this is the second one if the first one is wrong:

multiply 4.5 on both sides so that the 4.5/w would leave u with just only w alone.

then 3*4.5=13.5

w=13.5

hope dis helps if not then im not dat smart enough ;?

Here are a few fun facts:

A 19th century horse named 'Old Billy' is said to have lived 62 years.

Horses can sleep both lying down and standing up.

Horses can run shortly after birth.

Horses have around 205 bones in their skeleton.

Horses have been domesticated for over 5000 years.

Horses use their ears, eyes and nostrils to express their mood.

Because horse’s eyes are on the side of their head they are capable of seeing nearly 360 degrees at one time.

The fastest recorded sprinting speed of a horse was 88 kph (55 mph). Most gallop at around 44 kph or 27 mph.

The Przewalski’s horse is the only truly wild horse species still in existence. The only wild population is in Mongolia. There are however numerous populations across the world of feral horses e.g. mustangs in North America.

I hope you learned something new!

Answer:

Step-by-step explanation:

[tex]1\frac{1}{3} - 2\frac{1}{5} +\frac{7}{10}[/tex]

[tex]=-\frac{13}{15} +\frac{7}{10}[/tex]

[tex]=-\frac{1}{6}[/tex]

Step-by-step explanation:

H hvuvuvuvuvuv9yvig kg

Answer:

204 R45

Step-by-step explanation:

Answer: The answer will be 205 R2

Step-by-step explanation:

I know this because the steps to dividing are:

D - Divide

M - Multiply

S - Subtract

B - base(total)

If you do all of those steps then you'll get 205 R2

R = remainder(left overs)

So, the answer will be 205 R2

Step-by-step explanation:

1) 13.57

+2.93

With this, we subtract from the back Meaning we add starting from back to forward;

13.57

+2.93

16.50

We have that because when you add 7 and 3 you get 10 so we write the 0 there and add the remaining 1 to the 5 making it 6 and then 6 plus 9 will give you 15 then we write the 5 then we add the remaining 1 to the 3 making it 4 then we add the 4 to the 2 making it 6 after this there is no remainder so we just write the 1.

Answer:

no solution

Step-by-step explanation:

Answer:

54

Step-by-step explanation:

hi

Answer:

He can make 4 batches of sugar cookies.

Step-by-step explanation: 4x1/2=2 and because he lost 1 cup, he only has 2 cups left

Answer:

A. -5

Step-by-step explanation:

-9 + 4 = -5

hope this helps

45- 3, 5, 9

369- 3, 9

7870- 2, 5, 10

1976- 2, 4,

6003- 3, 9

136- 2, 3, 4

1674- 2, 3, 6, 9

35496- 2, 3, 4, 6, 9

Answer:

- 1.2+15.45=14.43

Step-by-step explanation:

Answer:

Step-by-step explanation:

The solution is, resulting matrix:[tex]\left[\begin{array}{ccc}4&-1&-4\\-4&6&2\\0&8&10\end{array}\right][/tex][tex]\left[\begin{array}{ccc}-7\\-4\\22\end{array}\right][/tex].

In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object. For example, is a matrix with two rows and three columns.

here, we have,

given that,

We are given a matrix,

Let A=[tex]\left[\begin{array}{ccc}4&-1&-4\\-4&6&2\\8&-4&6\end{array}\right][/tex]

and B=[tex]\left[\begin{array}{ccc}-7\\-4\\30\end{array}\right][/tex]

We have to find the resulting matrix after performing given row operations

2R2 +R3 ⇒ R3

Apply

we get,

:A=[tex]\left[\begin{array}{ccc}4&-1&-4\\-4&6&2\\0&8&10\end{array}\right][/tex]

and, B=[tex]\left[\begin{array}{ccc}-7\\-4\\22\end{array}\right][/tex]

Hence, resulting matrix

[tex]\left[\begin{array}{ccc}4&-1&-4\\-4&6&2\\0&8&10\end{array}\right][/tex][tex]\left[\begin{array}{ccc}-7\\-4\\22\end{array}\right][/tex]

To learn more on matrix click:

brainly.com/question/30646566

#SPJ7

Answer:

y = 2x -2

Step-by-step explanation:

When you shift down or up, you move down or up the y-axis. Here, the 6 was the y-intercept, so you subtract 8 from 6.

Answer:

After solving the compound inequality [tex]1.5x-1>6.5 \ or \ 7x+3<-25[/tex] we get [tex]\mathbf{x>5 \ or \ x<-4}[/tex]

Option B is correct.

Step-by-step explanation:

We are given compound inequality: [tex]1.5x-1>6.5 \ or \ 7x+3<-25[/tex]

We will first solve the inequalities to find the value of x, then will draw the graph.

[tex]1.5x-1>6.5 \ or \ 7x+3<-25\\1.5x-1+1>6.5+1 \ or \ 7x+3-3<-25-3\\1.5x>7.5 \ or \ 7x<-28\\x>\frac{7.5}{1.5} \ or \ x<\frac{-28}{7}\\x>5 \ or \ x< -4[/tex]

So, after solving the compound inequality [tex]1.5x-1>6.5 \ or \ 7x+3<-25[/tex] we get [tex]\mathbf{x>5 \ or \ x<-4}[/tex]

Now, the number line will be as shown in figure.

The description as given in options is:

A number line goes from negative 10 to positive 10. An open circle appears at negative 4 and positive 5. The number line is shaded from negative 4 toward negative 10. The number line is also shaded from positive 5 toward positive 10.

so, Option B is correct.

Answer:

the answer is b

Step-by-step explanation:

2020 edge

Answer: 6, but it isn't in your answer choices

Step-by-step explanation: When dividing exponents with the same base, subtract the bottom exponent from the top. In this case, you get 3 from subtracting n from 9, so n should be 6