need help with this geometry question please!

Answer:

yes

Step-by-step explanation:

the second number in a set of parenthesis is the y value and the number in the parenthesis is 92 which makes y=92 true

Answer:

[tex]\boxed{\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6} \bf x & -1 & 0&1&2&3\\\cline{1-6} \bf y & 10& 6& 2&(-2)&-6 \\\cline{1-6}\end{tabular}}[/tex]

Step-by-step explanation:

Given equation to us is :-

[tex]\implies y = 6 - 4x [/tex]

So , here we can put different values of x in order to get different values of y .

Initial table given to us is ,

[tex]\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6} \bf x & -1 & 0&1&2&3\\\cline{1-6} \bf y & & & 2&&-6 \\\cline{1-6}\end{tabular}[/tex]

Put x = (-1) :-

[tex]\implies y = 6 - 4x \\\\\implies y = 6 -4(-1) \\\\\implies y = 6 + 4 \\\\\red{\bf\implies y = 10} [/tex]

Put x = 0 :-

[tex]\implies y = 6 - 4x \\\\\implies y = 6-4(0) \\\\\implies y = 6 - 0 \\\\\pink{\bf\implies y = 6 }[/tex]

Put x = 2 :-

[tex]\implies y = 6-4x \\\\\implies y = 6 - 4(2) \\\\\implies y = 6 - 8 \\\\\blue{\bf\implies y = (-2)} [/tex]

Hence the final table would be :-

[tex]\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6} \bf x & -1 & 0&1&2&3\\\cline{1-6} \bf y & 10& 6& 2&(-2)&-6 \\\cline{1-6}\end{tabular}[/tex]

Answer:

Step-by-step explanation:

You answer goes to minus infinity. There is no answer and you are correct.

Answer:

Statement 1: Quadrilateral PQRT; line QSV; line PTV line QV bisects line RT, and line OR is parallel to line PV

Statement 3: angle QSR is congruent to angle VST

Statement 5: Triangle RSQ is congruent to angle TSV

Reason 1: Given

Reason 2: If a line bisects a segment, then it divides the segment into two congruent segments.

Reason 4: If two lines are parallel, then the alternate interior angles are congruent.

Step-by-step explanation:

The first statement will always be your given statements.

The given result is proved by ASA congruence criteria.

Two triangles are said to be congruent when all of their corresponding sides and angles are equal. For this relation between two triangles, there are many criteria such as SSS, SAS, ASA and RHS.

The given problem can be solved in statement and reason form as follows,

In triangles ΔQSR and ΔTSV,

Statement                                                           Reason

1. ∠QRS = ∠STV                                        Alternate interior angles for QR║PV

2. RS = TS                                                 QV bisects Line segment RT

3. ∠QRS = ∠VTS                                      Vertically opposite angles

4. ΔQSR ≅ ΔTSV                                     Triangles are congruent by ASA

Thus, QS =VS due to corresponding parts of congruent triangles.

Hence, QS = VS is proved by the ASA criteria for congruent triangles.

To know more about congruent triangles click on,

https://brainly.com/question/22062407

#SPJ2

Answer:

i) x = 2.8

ii) x = 2

Answer:

Situation B is a function.

Step-by-step explanation:

We know that,

A function is a relation in which every element in the domain i mapped to a unique element in the co-domain.

That is, every element in the domain must have a unique image.

So, we see that,

Situation A = {student's name, all the colors that the student likes}

Here, the students can have more than one color which they like.

Thus, this relation is not a function.

Situation B = {student's name, the student's favorite math teacher}

Here, the students can have only one favorite math teacher.

So, this relation is a function.

Answer:

D

Step-by-step explanation:

Answer: D) 13y^25 and 2y^25

Like terms involve the same variables, and each of those variables must have the same exponents.

Another example of a pair of like terms would be 5x^3y^2 and 7x^3y^2. Both involve the variable portion "x^3y^2" which we can replace with another variable, say the variable z. That means 5x^3y^2 becomes 5z and 7x^3y^2 becomes 7z. After getting to 5z and 7z, it becomes more clear we have like terms.

Answer:

=10

Step-by-step explanation:

Answer:

c) 33%

Step-by-step explanation:

In a  standard deck of cards, there are 52 cards total and of those 52, there are 12 face cards. One-third of those face cards are queens, (because there are 4 queens in a deck of cards, and [tex]\frac{4}{12}=\frac{1}{3}[/tex]). Therefore, there is a 33% chance that the card Luke holds is a queen.

Answer:

33%

Step-by-step explanation:

Got it right on the test.

Answer:

4

Step-by-step explanation:

7+7+7+7 or 28/7 =4

Answer:

I think the answer is 124 though I'm in 8th grade . I have completed algebra 1.

Step-by-step explanation:

The other side of the circle has four minor arcs since they each measure less than  180 .

Answer:

she babby sat atleast 4 hours last week

Step-by-step explanation:

.

The centroid is 2/3 of the opposite side. of the distance from each vertex to the midpoint. 8. To inscribe a circle about a triangle, you use the.

5 pages·2 MB

The complete statement is: the centroid is 2/3 of the distance from each vertex to the midpoint of the opposite side

The properties of a centroid are:

The second highlight above means that: the statement is to be completed with 2/3

Read more about centroids at:

https://brainly.com/question/1189196

Answer:

Fraction- 33/4

Decimal- 8.25000

Step-by-step explanation:

Hope this helps!

Brain-List?

Answer:

Average Mean = 1.8

Step-by-step explanation:

Missing:

X:0 1 2 3 4 5 6

P(x): 0.20 0.30 0.20 0.15 0.10 0.05

Computation:

Average Mean [Probability distribution]

∑[P(x) × X]

= 0 x 0.20 + 1 x 0.30 + 2 x 0.20 + 3 x 0.15 + 4 x 0.10 + 5 x 0.05

= 1.8

Average Mean = 1.8

Answer:

Step-by-step explanation:

yes

if you solve the equation,

5x+15=7x-11

26=2x

x=26/2=13

Answer:

Step-by-step explanation:

There are total cadies in the bag:

Probability of pulling a coffee candy is:

So each 5th candy should be coffee candy

Then you need to pull at least:

Note:

To ensure 3 coffee candies you would need to pull 25 - 2 = 23 candies

Answer:

y = 2x +4

Step-by-step explanation:

m = [tex]\frac{y_{2} -y_{1} }{x_{2} - x_{1} }[/tex]

y - [tex]y_{1}[/tex] = m ( x - [tex]x_{1}[/tex] )

~~~~~~~

(1, 6)

(- 3, - 2)

m = 2

y - 6 = 2 ( x - 1 ) <------ ( point-slope form )

y = 2x + 4  <------- ( slope-intercept form )

2x - y = - 4  <------ ( standard form )

Step-by-step explanation:

Let the given ratio be pk : qk .

So , here the quadratic equation is lx² + nx + n = 0. With respect to Standard form ax² + bx + c = 0.

We have ,

→ Sum of roots = -b/a = -n/l = qk + pk

→ Product of roots = c/a = n/l = k²pq .

[tex]=> \dfrac{n}{l} = \dfrac{k^2}{pq} \\\\=> k^2 =\dfrac{n}{pql} [/tex]

And here pk and qk is a root of the quadratic equation ,

[tex]=> lx^2 + nx + n = 0 \\\\=> l(pk)^2 + n(pk) + n = 0\\\\=> lp^2k^2+npk + n = 0 \\\\=> lp^2\bigg( \dfrac{n}{pql} \bigg) + np\bigg(\sqrt{\dfrac{n}{pql}} \bigg) + n = 0 \\\\ => n\bigg\{\dfrac{p}{q}+\sqrt{\dfrac{np}{lq}}+1\bigg\} = 0 \\\\=> \dfrac{p}{q}+\sqrt{\dfrac{np}{lq}}+1 =0\\\\=>\sqrt{\dfrac{p}{q}} \bigg( \dfrac{q}{p}+\sqrt{\dfrac{np}{lq}}+1\bigg) = 0 \\\\=> \sqrt{\dfrac{p}{q}}+ \sqrt{\dfrac{q}{p}}+ \sqrt{\dfrac{n}{l}}=0 \\\\\boxed{\red{\bf\longmapsto \sqrt{\dfrac{p}{q}}+ \sqrt{\dfrac{q}{p}} = - \sqrt{\dfrac{n}{l}}}} [/tex]

Answer:1.5

Step-by-step explanation:

Answer:

endpoint for B is (10,-2)

Step-by-step explanation:

M={(x1+x2)/2,(y1+y2)/2}

(4,2)={-2+x2/2,6+y2/2)

By equating the x coordinates

4=-2+x2/2

8=-2+x2

8+2=x2

x2=10

By equating the y coordinates

2=6+y2/2

4=6+y2

4-6=y2

y2=-2

Therefore the endpoint for B is (10,-2)

Answer:

.06

Step-by-step explanation:

edg

Answer:

.06

Step-by-step explanation:

test on edg e

Answer:

4

Step-by-step explanation:

y = mx + b

m represents the slope, and as we can see in the equation, the number 4 is there instead of m, so therefore 4 is the slope

Answer:

x = -2/-3

Step-by-step explanation:

since you have your variable on one side and your whole number on the other, divide by -3 on both sides to get x=-2/-3

Answer:

A) The point( -1 , 5)  is satisfies the  given graph  [tex]g(x) = (\frac{1}{5} )^{x}[/tex]

B) The point( 3 , 1/125)  is satisfies the  given graph  [tex]g(x) = (\frac{1}{5} )^{x}[/tex]

Step-by-step explanation:

Explanation:-

Given graph  

               y =      [tex]g(x) = (\frac{1}{5} )^{x}[/tex]

Put the point ( -1 , 5)

Put y =5 and x =-1

            [tex]5 = (\frac{1}{5} )^{-1} = 5[/tex]

The point( -1 , 5)  is satisfies the  given graph

ii)

Given graph  

               y =      [tex]g(x) = (\frac{1}{5} )^{x}[/tex]

             [tex]\frac{1}{125} = (\frac{1}{5} )^{3} = \frac{1}{125}[/tex]

The point( 3 , 1/125)  is satisfies the  given graph  [tex]g(x) = (\frac{1}{5} )^{x}[/tex]

Answer:

f(x)=-3x+9

Step-by-step explanation:

Answer:

option A) Agree, because the solutions are the same is correct.

Step-by-step explanation:

FIRST SYSTEM

[tex]6x + y= 2[/tex]

[tex]-x-y=-3[/tex]

solving the system

[tex]\begin{bmatrix}6x+y=2\\ -x-y=-3\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18[/tex]

[tex]\begin{bmatrix}6x+y=2\\ -6x-6y=-18\end{bmatrix}[/tex]

adding the equation

[tex]-6x-6y=-18[/tex]

[tex]+[/tex]

[tex]\underline{6x+y=2}[/tex]

[tex]-5y=-16[/tex]

so the system becomes

[tex]\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}[/tex]

solve -5y for y

[tex]-5y=-16[/tex]

Divide both sides by -5

[tex]\frac{-5y}{-5}=\frac{-16}{-5}[/tex]

simplify

[tex]y=\frac{16}{5}[/tex]

[tex]\mathrm{For\:}6x+y=2\mathrm{\:plug\:in\:}y=\frac{16}{5}[/tex]

[tex]6x+\frac{16}{5}=2[/tex]

subtract 16/5 from both sides

[tex]6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}[/tex]

[tex]6x=-\frac{6}{5}[/tex]

Divide both sides by 6

[tex]\frac{6x}{6}=\frac{-\frac{6}{5}}{6}[/tex]

[tex]x=-\frac{1}{5}[/tex]

Therefore, the solution to the FIRST SYSTEM is:

[tex]x=-\frac{1}{5},\:y=\frac{16}{5}[/tex]

SECOND SYSTEM

[tex]2x-3y = -10[/tex]

[tex]-x-y=-3[/tex]

solving the system

[tex]\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6[/tex]

[tex]\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}[/tex]

[tex]-2x-2y=-6[/tex]

[tex]+[/tex]

[tex]\underline{2x-3y=-10}[/tex]

[tex]-5y=-16[/tex]

so the system of equations becomes

[tex]\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}[/tex]

solve -5y for y

[tex]-5y=-16[/tex]

Divide both sides by -5

[tex]\frac{-5y}{-5}=\frac{-16}{-5}[/tex]

Simplify

[tex]y=\frac{16}{5}[/tex]

[tex]\mathrm{For\:}2x-3y=-10\mathrm{\:plug\:in\:}y=\frac{16}{5}[/tex]

[tex]2x-3\cdot \frac{16}{5}=-10[/tex]

[tex]2x=-\frac{2}{5}[/tex]

Divide both sides by 2

[tex]\frac{2x}{2}=\frac{-\frac{2}{5}}{2}[/tex]

Simplify

[tex]x=-\frac{1}{5}[/tex]

Therefore, the solution to the SECOND SYSTEM is:

[tex]x=-\frac{1}{5},\:y=\frac{16}{5}[/tex]

Conclusion:

As both systems of equations have the same solution.

Therefore, we conclude that Albert is right when says that the two systems of equations shown have the same solutions.

Hence, option A) Agree, because the solutions are the same is correct.

Answer:

It's pretty simple. For example, if it says there is a percentage increase on something like there is a jacket for $25 and there was a 15% increase, you multiply 25 by 1.15 to see what the new price is. But say if the same $25 jacket is on sale for 15%, then you can multiply 25 by .15, then subtract that from 25 or originally subtract the 15% from 100 which is 85 and multiply 25x.85 and get the answer directly.

Hope this helps! But this would be easier to explain with an example