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2 years ago

Please answer this question

Mathematics

1 answer:

lana66690 [7]2 years ago

8 0

Answer:

b (0, 4)

Step-by-step explanation:

The other options already have an input (the x) in the function so it must be b.

Las otras opciones ya tienen una entrada (la x) en la función, por lo que debe ser b.

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10y - 3y = 15 graph using the table of values

harina [27]

Answer:

y+15/7

Step-by-step explanation:

i took this test

6 0

3 years ago

Points b d and f are midpoints of ten sides of ace ec=44 df=25 find ac

scZoUnD [109]

Answer:

The length of AC is;

C. 50

Step-by-step explanation:

By the midpoint of a triangle theorem, we have that a segment that spans across and intersects with the midpoints of two sides of a triangle is equal to half the length of the third side and parallel to the length of the third side

The given parameters are;

The midpoints of ΔACE are B, D, and F

The length of EC = 44

The length of DF = 25

Therefore, we have;

Given that DF is a midsegment of triangle ΔACE, then DF ║ AC and

the length of DF = (1/2) × AC the length of AC

∴ The length of AC = 2 × The length of DF

The length of DF = 25

∴ The length of AC = 2 × 25 = 50

The length of AC = 50

4 0

2 years ago

Plz helpppp I don’t GET THIS

Liono4ka [1.6K]

First you have to find 1/3 of the marbles. That means in the twelve marvels she has, there are three even groups. Divide twelve by three and you get four. So if each group is four, it is simple from here on out. If she started out with twelve and she lost 1/3 of them(4) it would be 12-4. You know this because lost is another for for subtract. 12-4=8. She has 8 marbles left.

6 0

2 years ago

Read 2 more answers

Brainiest if correct or first<br> 1/2(6x+1/2)

Alexxx [7]

Step-by-step explanation:

1/2(6x+1/2)=0

3x+1/4=0

3X=-1/4

X=3(-1/4)

X=-3/4

X=-.75

8 0

2 years ago

Read 2 more answers

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago