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2 years ago

Calculate the volume of the figure.

Mathematics

2 answers:

Margarita [4]2 years ago

8 0

Answer:

Step-by-step explanation:

Formula

Volume = L * w * h

Givens

L = 7 feet

w = 4 feet

h = 5 feet

Solution

V = L * w * h Substitute values into the formula

V = 7 * 4 * 5

V = 140 cubic feet.

Answer

140 ft^3

Effectus [21]2 years ago

6 0

Answer:

D) 140 cubic feet

Step-by-step explanation:

The given information is,

→ Length (l) = 7 ft

→ Breadth (b) = 4 ft

→ Height (h) = 5 ft

Formula we use,

→ Volume of cuboid = l × b × h

Let's solve the problem,

→ l × b × h

→ 7 × 4 × 5

→ [ 140 ft³ ]

Hence, the volume is 140 ft³.

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Part 2 - Find the error(s) and solve the problem correctly.

Aneli [31]

Answer:

Step-by-step explanation:

6 0

2 years ago

12x−5y=−20 y=x+4 x=x=x, equals y=y=y, equals

aksik [14]

Answer:

The solution is (0, 4)

Step-by-step explanation:

Please pay attention to the first two equations and drop the last two:

12x−5y=−20 y=x+4 x=x=x, equals y=y=y should ideally be:

12x−5y=−20

y=x+4

Let's find x. Substitute x + 4 for y in the first equation, obtaining:

12x - 5(x + 4) = -20

Carrying out the indicated multiplication, we get:

12x - 5x - 20 = -20, or 7x = 0

If x = 0 then y must be 0 + 4, or 4.

The solution is (0, 4)

8 0

3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

5 0

2 years ago

Virginia has a monthly budget of $1,900. If Virginia earns $11.00 per hour, and works 145 hours this month, will Virginia earn e

mart [117]

Answer:

no she will not

Step-by-step explanation:

145*11 < 1900

6 0

2 years ago

Read 2 more answers

HELP!!!!

ch4aika [34]

Answer:

$8 y^{50}$

Step-by-step explanation:

Given (64 y Superscript 100 Baseline) Superscript one-half.

Let us write it into an equation.

$\left(64 y^{100}\right)^{\frac{1}{2}}$

Apply radical rule: $\sqrt[n]{a}=a^{\frac{1}{2}}$ and $a^{m+n}=a^m+a^n$

$\begin{aligned}\left(64 y^{100}\right)^{\frac{1}{2}} &=\sqrt[2]{64 y^{100}} \\&=\sqrt[2]{8^{2} y^{50} y^{50}} \\&=\sqrt[2]{8^{2}\left(y^{50}\right)^{2}} \\&=8 y^{50}\end{aligned}$

Hence, $8 y^{50}$ is equivalent to (64 y Superscript 100 Baseline) Superscript one-half.

8 0

3 years ago