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4 years ago

For the function f(x) = x^2+ 3, what are the outputs for the inputs -3, 2, 0, and 5?

Mathematics

1 answer:

AysviL [449]4 years ago

8 0

Answer:

F(-3)=12

F(2)=7

F(0)=3

F(5)=28

Step-by-step explanation:

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Let f(x)=x^2+7 and(x)=x+3/x. find (g*f)(-5)

Karolina [17]

Hello hereb is a solution :

5 0

4 years ago

the primiter of a triangle is 110cm. side x of the triangle is four times as long as side y. side z is 2cm longer side x. find t

Alex787 [66]

Answer:

x is 48 y is 12 z=50

Step-by-step explanation:

x+y+z=110

x=4y

4y+2=z

4y+y+4y+2=110

9y+2=110

9y=108

y=12

x=48

z=50

Hope this helps!

7 0

3 years ago

A cylinder shaped juice pitcher has a diameter of 12 cm and a height of 25 cm. What volume of juice does the pitcher contain whe

Andreas93 [3]

Answer:

706.50 cubic cm.

Step-by-step explanation:

Given that a cylinder shaped juice pitcher has a diameter of 12 cm and a height of 25 cm.

We are to find out the t volume of juice does the pitcher contain when it is 25% full. 3.14 is to be used for pi.

Volume of the cylinder = $\pi r^2 h\\ \\=3.14 *6^2*25\\=2826$

When it is 25% full the volume would be 1/4 of the full volume

= $\frac{2826}{4} =706.50$ cubic cm

So volume when 25% full is

706.50 cubic cm.

4 0

4 years ago

Simplify: 1/3x(6x - 1) A) 5 3 x B)2x - 1 C)2x - 1 3 D)2x2 - 1 3 x

Nina [5.8K]

Multiplying out the expression, you get
2x^2 -(1/3)x . . . . . . matches selection D

5 0

4 years ago

Read 2 more answers

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean $\mu =266\\$

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find $P(X\geq 308)$

Using normal distribution,

$z=\frac{X-\mu}{\sigma}$

$z=\frac{308-266}{15}$

$=\frac{42}{15}$

Thus $z=2.8$

So $P\left (X\geq 308 \right )=1-P(X$

$=1-P(z$

$=1-Table\: value\: of\: 2.8$

$=1-0.99744$

$=0.00256$

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3%.

The z-value for the lowest of 3% is -1.8808

Then $X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu$

This implies $X=-1.8808*15+266=237.788$

Thus the babies who are born on or before 238 days are considered to be premature.

5 0

3 years ago