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2 years ago

7

Help please, IT'S SUPER DUPER EASY

2 answers:

ki77a [65]2 years ago

7 0

Answer:

3/16 hope this helps

V125BC [204]2 years ago

5 0

The answer would be 18/96. Hope this helps and it was ez.

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The attendance at two baseball games on successive nights was 74,000. The attendance on Thursday's game was 4,000 more than two-

nadezda [96]

Answer:

The answer is 53,333 rounded

Step-by-step explanation:

So you multiply 74,000 by 2/3 you get 49,333

Then you take that number and add 4,000 and you get 53,333 rounded.

6 0

3 years ago

Select all the descriptions that<br> apply to the transformation.

Pavel [41]

Answer:

B and D

Step-by-step explanation:

As we see the transformation involves translation left and down by 6 units

<u>Correct answer choices are:</u>

B. 6 units down, 6 units left
D. T(-6, -6)

4 0

2 years ago

The hypotenuse of a right triangle measures 20 cm and one of its

Ket [755]

I’m pretty sure 17.3 because of you do a^2+b^2=c^2 you get a^2= 10 square root 3 and that in a decimal is 17.3205

5 0

2 years ago

Read 2 more answers

The circumference of this tabletop is 24π inches, Which of the following best describes how to determine the area of the tableto

omeli [17]

Answer:

D Find half of 24 to calculate the radius. Then square the radius and multiply the result by π.

Step-by-step explanation:

Given that circumference of circle is: $24 \pi$

The formula for circumference of circle is:

$C = 2\pi r$

Comparing both formulas, we get

$24\pi = 2\pi r\\24 = 2r\\r = \frac{24}{2}\\r = 12$

The radius is obtained by dividing 24 by 2.

We also know that the area of circle is:

$A = \pi r^2$

The radius is being squared and then multiplied by pi to get the area.

Hence,

The correct answer is:

D Find half of 24 to calculate the radius. Then square the radius and multiply the result by π.

3 0

2 years ago

Find an n-degree polynomial function with real coefficients satisfying the given condition.

kolezko [41]

In all cases, if $f$ has real coefficients, then any complex roots occur in conjugate pairs, so if $a+bi$ is a root, then so is $a-bi$ . Also, by the fundamental theorem of algebra, if $r_1,\ldots,r_n$ are roots to $f$ , then for some constant $a\in\mathbb R$ ,

$f(x)=a(x-r_1)\cdots(x-r_n)$

1. If $n=3$ and $f(3)=f(2i)=0$ , then

$f(x)=a(x-3)(x-2i)(x+2i)=ax^3-3ax^2+4ax-12a$

Given that $f(-1)=50$ , we have

$f(-1)=a(-1-3)(-1-2i)(-1+2i)=-20a=50\implies a=-\dfrac52$

$\implies\boxed{f(x)=-\dfrac52x^3+\dfrac{15}2x^2-10x+30}$

2.

$f(x)=a(x-4)(x-(-5+2i))(x-(-5-2i))=a x^3 + 6 a x^2 - 11 a x - 116 a$

With $f(2)=-636$ , we have

$f(2)=a(2-4)(2+5-2i)(2+5+2i)=-106a=-636\implies a=6$

$\implies\boxed{f(x)=6x^3+36x^2-66x-696}$

The rest are done in the same exact way.

3 0

2 years ago