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soldi70 [24.7K]

3 years ago

9

A toy chest wall hole 78 ft.³ of toys the height of the chest is 4 feet and the width is 3 feet what is the length of the toy ch

est

1 answer:

Anit [1.1K]3 years ago

5 0

The answer is 6.5 because if you multiply 4 and 3, you only need to find what you need to multiply to get to the goal number, 78.

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What is 95x9 divided by 14?

SIZIF [17.4K]

Answer:

213.75

Step-by-step explanation:

95×9=855

855÷4=213.75

3 0

1 year ago

Please help me thanks if u do

aniked [119]

Answer: a

Step-by-step explanation: look at the chart

3 0

3 years ago

Is 6.3 > 6.04 reasonable yes or no Explain

Brums [2.3K]

Answer:

Yes

Step-by-step explanation:

By rounding to the 10ths place, we can easily see that 6.3 is greater than 6.04. 6.3 is already rounded to the 10ths place, but 6.04 rounded to the 10ths place is 6.0.

6.3 is clearly more than 6.0, therefore 6.3 is greater than 6.04.

3 0

2 years ago

A pile of earth removed from an excavation is a cone measuring 6ft high and 30ft across its base. How many trips will it take to

nignag [31]

Answer: 14

Step-by-step explanation:

Given

Excavation cone measures

height $h=6\ ft$

Diameter $d=30\ ft$

Truck can dump $125\ ft^3$ at a time

The volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$

Putting values

$\Rightarrow V=\dfrac{1}{3}\times \pi\times 15^2\times 6\\\Rightarrow V=1413.9\approx 1414\ ft^3$

No of trips(N) required to accumulate this much volume is given by

$\Rightarrow N=\dfrac{1414}{125}=13.3\approx 14$

Therefore, 14 trips are necessary to accumulate a cone of volume $1414\ ft^3$

8 0

2 years ago

Trig proofs with Pythagorean Identities.

lorasvet [3.4K]

To prove:

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$ .

$$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}$

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

$$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}$

Denominators are same, you can subtract the fractions.

$$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}$

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

$$= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

$$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

$$=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x}$ ------------ (1)

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

$$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

$$=1+2\frac{\cos^{2} x}{\sin^{2} x}$

$$=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x}$ ------------ (2)

Equation (1) = Equation (2)

LHS = RHS

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.

5 0

3 years ago