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3 years ago

Help :( someone can explain me

Mathematics

1 answer:

Inessa [10]3 years ago

6 0

Answer:

B, C, D

Step-by-step explanation:

-2 is 8 spaces away from 6 so do the math for each possible answer and u get B,C,D.

6 - (-2)

6 + 2 = 8

| -2| = 2

6+2 = 8

|-2+6|

2+6 = 8

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Evaluate the function f(x) = x^2 +3x + 5 for x= -2

forsale [732]

Answer:

f(-2) = 3

Step-by-step explanation:

f(x) = x² + 3x +5

f(-2)=(-2)² + 3*(-2)+5

= 4 - 6 +5

= 3

6 0

4 years ago

How do you solve this using substitution? <br> -x+y=1<br> x+y=-1

Alexeev081 [22]

Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point Form: (-1,0)
Equation form: x=-1,y=0

4 0

4 years ago

PLZ ANSWER PLZ PLZ PLZ HELP 1 QUESTION!!!

JulsSmile [24]

We first must find the value of angle z.

Angle z = 90° - 42°

Angle z = 48°

We now use the sine function to find y.

sin(48°) = 35/y

y = 35/sin(48°)

y = 47.0971455362

We now round off to two decimal places.

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I hope this helps....

8 0

3 years ago

What letter follows next in this sequence?<br> A, d, i, p.

vivado [14]

A to D - 2 letters in between

D to I - 4 letters in between

I to P - 6 letters in between

P to Y - 8 letters in between

the answer should be Y.
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4 0

3 years ago

Read 2 more answers

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).

4 0

3 years ago