Im confused someone solve for me! helppp

Answer these 3 pls find the shaded region

murzikaleks [220]

7) 100in²

8) 56cm²

9) 51in²

Have a great day <3

5 0

3 years ago

Read 2 more answers

We are going to fence in a rectangular field. If we look at the field from above the cost of the vertical sides are $10/f t, the

Brut [27]

Answer:

Dimensions are 350/9 ft and 17.5 ft

Step-by-step explanation:

We are given the cost per ft of all the 4 sides. Let the horizontal be x and the vertical be y.

Now, we will set up the constraint and equation that we are being asked to maximize.

Thus;

700 = 10y + 10y + 7x + 2x

700 = 20y + 9x

Maiking y the subject, we have;

y = (700 - 9x)/20

y = 35 - 9x/20

Now,area of a rectangle is: A = xy

Thus, A = x(35 - 9x/20))

A = 35x - 9x²/20

We can get the critical points by finding the derivatives and Equating to zero

Thus;

dA/dx = 35 - 0.9x

At dA/dx = 0,we have; x = 350/9

At d²A/dx², we have;

d²A/dx² = -0.9

This is negative, thus we will disregard and use the one gotten from the first derivative.

Thus, we will use x = 350/9 ft

Plugging this into the equation y = 35 - 9x/20, we have;

y = 35 - ((9 × 350/9)/20)

y = 17.5 ft

7 0

4 years ago

If TV ≅ WV and UV ≅ XV, then which congruency theorem shows that ΔTUV ≅ ΔWXV ?

VashaNatasha [74]

Answer:

Option (2).

Step-by-step explanation:

In the figure attached,

Given : TV ≅ WV and UV ≅ XV

To prove: ΔTUV ≅ ΔWXV

Statements Reasons

1). TV ≅ WV 1). Given

2). UV ≅ XV 2). Given

3). ∠XVW ≅ ∠TVU 3). Reflexive property

4). ΔTUV ≅ ΔWXV 4). SAS (Side - angle - side) property of congruence

Therefore, Option (2) will be the answer.

5 0

4 years ago

Datguy323 is going to complain again. What's the variables for: <img src="https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D29%5C%5Cx%2By%

Korolek [52]

Answer: :o I FINALLY MADE IT

(5, 2)

x = 5

y = 2

Step-by-step explanation:

First, I graphed both equations. They meet at the points (5,2) and (2,5). Because y < 5, the solution is (5, 2)

<em>Hope it helps <3</em>

8 0

3 years ago

Read 2 more answers

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago