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3 years ago

7

A square has an area of 16x^2 square inches. If the sides of the square were to increase by 2 units, what is the new area of the

square?

1 answer:

kykrilka [37]3 years ago

4 0

Answer:

16x² + 64x + 64

Step-by-step explanation:

Given

A = 16x²

Increasing the sides by 2, substitute x = x + 2 into A

A = 16(x + 2)² ← expand factor using FOIL

= 16(x² + 4x + 4 ) ← distribute

= 16x² + 64x + 64

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Please help me out on this one !!

Serjik [45]

I think the answer is c

5 0

3 years ago

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine foun

QveST [7]

Answer:

We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 31.7 milligrams

Sample mean, $\bar{x}$ = 28.5 milligrams

Sample size, n = 9

Alpha, α = 0.05

Sample standard deviation, s = 2.8 milligrams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 31.7\text{ milligrams}\\H_A: \mu < 31.7\text{ milligrams}$

We use One-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{28.5 - 31.7}{\frac{2.8}{\sqrt{9}} } = -3.429$

Now, $t_{critical} \text{ at 0.05 level of significance, 8 degree of freedom } = -1.860$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and accept the alternate hypothesis. We conclude that the mean nicotine content is less than 31.7 milligrams for this brand of cigarette.

3 0

3 years ago

10. Which of the following corresponds to the midpoint of the points A(-2, 5, -3) and B(7,-4, 10)

pochemuha

Answer:

Step-by-step explanation:

$mid ~point~of~A(-2,5,-3)~and~B(7,-4,10) ~is~C(\frac{-2+7}{2} ,\frac{5-4}{2} ,\frac{-3+10}{2} )\\=(\frac{5}{2},\frac{1}{2} ,\frac{7}{2} )$

4 0

3 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

3 years ago

Please Help!!!! Find two consecutive odd numbers whose product is 47 more than their sum.

fenix001 [56]

7 and 9

7 x 9 = 63

7+9=16

63-16 = 47

7 0

2 years ago