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3 years ago

If point F is the midpoint of segment DE, which statement is true?

1 answer:

5 0

Segment DF is 1/2 the length of Segment DE. Since the entire segment is DE and it's midpoint is at F, the segment would be broken into 2 equal segments in one. Segment DF and FE. Segment DF is congruent to Segment FE.

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One more than the product of 15 and g

givi [52]

(15 x g) + 1

is your answer

hope this helps

7 0

3 years ago

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For a school fundraiser, Ayana needs to sell 40 rolls of wrapping paper. So far, she has sold 7 rolls to her grandmother, 2 roll

ICE Princess25 [194]

Answer:

22?

Step-by-step explanation:

Just take 40 and subtract the 7 to get 33 and then 2 to get 31 and 9 to get 22.

4 0

3 years ago

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Will give brainliest, thanks, and 5 stars! LOTS OF POINTS! 50 POINTS!

sweet-ann [11.9K]

Answer/Step-by-step explanation:

All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2

Answer: False

Explanation:

$3x^2-12+12=0+12$

$3x^2=12$

$\frac{3x^2}{3}=\frac{12}{3}$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two unique solutions to the equations (x-3)^2 = 16

Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.

Answer: True

Explanation:

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0

Answer: False

Explanation:

$2\left(x-3\right)^3-18+18=0+18$

$2\left(x-3\right)^3=18$

$\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}$

$\left(x-3\right)^3=9$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Answer: False

Explanation:

$2\left(x-5\right)^2-8+8=0+8$

$2\left(x-5\right)^2=8$

$\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}$

$\left(x-5\right)^2=4$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

Answer: False

Explanation:

$\left(x+3\right)^2-25+25=-8+25$

$\left(x+3\right)^2=17$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{17}-3,\:x=-\sqrt{17}-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

Answer: False

Explanation:

$\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}$

$\left(2x-1\right)^2=9$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=2,x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The only solution for equation (2x-1)^2-49=0 is x = 4

Answer: False
Explanation:

$\left(2x-1\right)^2-49+49=0+49$

$\left(2x-1\right)^2=49$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=4,\:x=-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

Answer: True
Explanation:

$3\left(x+2\right)^2-3+3=0+3$

$3\left(x+2\right)^2=3$

$\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}$

$\left(x+2\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=-1,\:x=-3$