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geniusboy [140]

3 years ago

15

I need help with number 4

1 answer:

Temka [501]3 years ago

7 0

Hello!

We see that c is less than 27.5. If we have c on the right, we will use <.

Therefore, our answer is B) c<27.50

I hope this helps!

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4 less than the product of 8 and a number is equal to 6

ICE Princess25 [194]

5 I think that should

4 0

3 years ago

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Which of these describes the unit rate?

AlexFokin [52]

Answer:

Which of these describes the unit rate? answer choices. a unit of measure. a fraction with a numerator of one. a fraction with a denominator of one. a proportion.

Step-by-step explanation:

Hope this helps you!

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5 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> What is [100(−4)^n-1} equal to?

nikklg [1K]

Answer: -5100

<u>Step-by-step explanation:</u>

$\sum^4_1[100(-4)^{n-1}]\qquad \rightarrow \qquad a_1=100\ \text{and r = -4}\\\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\\\\\S_4=\dfrac{100(1-(-4)^4)}{1-(-4)}\\\\\\.\quad=\dfrac{100(1-256)}{1+4}\\\\\\.\quad=\dfrac{100(-255)}{5}\\\\.\quad=20(-255)\\\\.\quad=-5100\\$

3 0

2 years ago

Draw the graph of 3x+2y+6

Elanso [62]

Answer:

cymath.com

Step-by-step explanation:

you can solve other equations there as well

7 0

2 years ago

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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

2 years ago