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4 years ago

14

X 9 =12 2 ⋅12 7 . x=?

1 answer:

ValentinkaMS [17]4 years ago

7 0

I think it’s
x= -54
PLZZ GIVE BRAINLYEST

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PLZ HELP!!!<br><br>What is 9 + 10?

vovangra [49]

The answer is 19 lol

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3 years ago

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Write the equation for this line in slope-intercept form.

Naya [18.7K]

Answer:

The answer is A. -4.

Step-by-step explanation:

The missing value you are looking for is the y-intercept or where the line crosses the y axis.

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3 years ago

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At to-go pizza each pizza cost $12 and each small salad costs $2. A delivery charge of $3 is added to every order. How much woul

MariettaO [177]

First, let's make some variables then make an expression.

Let x = the number of pizzas
Let y = the number of salads

Then 12x = the price for x pizzas
Then 2y = the price y pizzas

Now we can make our expression.

We combine the costs for the pizzas and salads and delivery charge is added.

12x + 2y + 3

Now we have our expression. Replace the x with how many pizzas you bought and replace the y with how many salads you bought.

How much would 6 pizzas and 3 small salads would cost?

12(6) + 2(3) + 3 =
72 + 6 + 3 =
78 + 3 =
81

The answer to this question is $81.

EDIT:

I want to give a thank you to Irspow for finding my error!

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3 years ago

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Can you help me to find the answer

const2013 [10]

He simplified expression C because that is the only one that has only multiplications.

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3 years ago

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

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3 years ago