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4 years ago

Consider the isosceles triangle with verticles a(-1,1) b(3,1) c(1,6)

Mathematics

1 answer:

Dima020 [189]4 years ago

8 0

What is the question? Do they want us to find a’,b’, and c’??

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On of Tho Mcd 8. Write the standard form of the equation y + 4 = - 12 (x – 1).

LiRa [457]

Answer:

12x + y = 16

Step-by-step explanation:

Standard Form: Ax + By = C

Step 1: Write point-slope form

y + 4 = -12(x - 1)

Step 2: Find slope-intercept form

y + 4 = -12x + 12

y = -12x + 16

Step 3: Find standard form

12x + y = 16

5 0

3 years ago

Use a common denominator to determine whether the ratios form a proportion.

scoray [572]

B is the correct answer. you can not multiply the first ratio be anything to make it equal the 2nd.

6 0

3 years ago

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The Spanish Club has 78 members, 42 girls and 36 boys. What is the ratio of

elena-14-01-66 [18.8K]

Answer:

C. 7:6

Step-by-step explanation:

Take your ratio 42 (G): 36 (B) and divide both numbers by six. This comes out to 7 (G): 6 (B) which is the simplest form of this ratio.

5 0

3 years ago

Read 2 more answers

Help me please!!!!!!!

lilavasa [31]

I think the first one is C

8 0

3 years ago

Mai wants to make an open top box by cutting out corners of a square piece of cardboard and folding up the sides. The cardboard

Bumek [7]

Answer:

$V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$

The domain for x is all real numbers greater than zero and less than 5 com

Step-by-step explanation:

The question is

What is the volume of the open top box as a function of the side length x in cm of the square cutouts?

see the attached figure to better understand the problem

Let

x -----> the side length in cm of the square cutouts

we know that

The volume of the open top box is

$V=LWH$

we have

$L=(10-2x)\ cm$

$W=(10-2x)\ cm$

$H=x)\ cm$

substitute

$V(x)=(10-2x)(10-2x)x\\\\V(x)=(100-40x+4x^{2})x\\\\V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$

Find the domain for x

we know that

$(10-2x) > 0\\10> 2x\\ 5 > x\\x < 5\ cm$

The domain is the interval (0,5)

The domain is all real numbers greater than zero and less than 5 cm

therefore

The volume of the open top box as a function of the side length x in cm of the square cutouts is

$V(x)=(4x^{3}-40x^{2}+100x)\ cm^3$

5 0

3 years ago