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2 years ago

Round the fraction 7 4/5 to the nearest whole number

Mathematics

2 answers:

MatroZZZ [7]2 years ago

8 0

The correct answer is: 8

Explanation:

Given mixed number:

$7\frac{4}{5}$

In order to convert a mixed number into a decimal number, follow the steps mentioned below:

Multiple 7 with 5 and add the numerator to the answer you get by multiplying 7 with 5. After that, write that result in the numerator and divide it with 5:

Step-1:

Multiply 7 with 5.

7*5 = 35

Step-2:

Add numerator value, which is 4 in this case, in 35

35 + 4 = 39

Step-3:

Rewrite the fraction.

$\frac{39}{5}$

Step-4:

Calculate.

$\frac{39}{5} = 7.8$

Now, as the number at the tenth place after the decimal point is greater than 5, we have to round it off to 8.0.

Hence the correct answer is 8.

Over [174]2 years ago

3 0

Im not sure if you are saying 74/5 or 7 and 4 fifths but in the case you'r saying 74/5 you would first round to see the highest number that 5 can go to without being a decimal. Which would be 70 so you would count how many times 5 can go into 70, which would be 14, so the left over is 4/5 so you would check to see if it can be simplified which it can't so the final answer is 14 and 4/5

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What is the width of a rectangle with length 14cm and area 161 cm^2

Inessa05 [86]

Answer:

Step-by-step explanation:

The area of a rectangle can be found with the formula $A = l*w$ , where $l$ is the length of the rectangle and $w$ is the width.

From the given problem statement, we know that $l = 14$ and $A = 161$ , so we can fill in those values in the formula anbd solve for $w$ to get the width:

$A = l*w$

$161 = 14w$

Divide both sides by $14$ to isolate $w$ by itself:

$\frac{161}{14} = \frac{14w}{14}$

$w = 11.5$

3 0

2 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago

Part of the area under this ramp serves for storage. Items can be stored only in the unshaded part of the ramp

Mekhanik [1.2K]

Answer:

48.1 m³ available, 2.5m³ not available

Step-by-step explanation:

Tan(X) = 4.5/(5.8+1.7)

Tan X = 3/5

Hypotenuse² = 3² + 5² = 34

Sin X = 3/sqrt(34)

X = 30.96

1/sinX = 1.7/SinY

Y = 61.003°

Third angle: 180 - X - Y = 88.037

Volume

= ½ × 1.7 × 1 × sin(88.038) × 3

= .

= 2.548

= 2.5m³

Available for storage:

½×4.5×7.5×3 - 2.548

= 48.077 = 48.1 m³

7 0

2 years ago

Math escape room maze ( I need help) I know it goes A,E,I but I don’t know what goes next)

miskamm [114]

Answer:

Step-by-step explanation:

Use angle summation:

∠HGE + ∠EGF = ∠HGF

3x + 11 + 110 = 16x + 4

117 = 13x

x = 9

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2 years ago

In the figure, m and n are rays.

blagie [28]

Answer:

A is correct because 1 and 3 are vertical angles.

B is correct because there are 180 degrees in a triangle. 180-65-55 = 60. Angle 2 is vertical to the third angle, so angle 2 equals 60.

C is incorrect because it is an exterior angle. Exterior angles are equal to the sum of the two isolated interior angles. 65 + 55 ≠ 60.

D is correct because 3 is an exterior angle. Exterior angles are equal to the sum of the two isolated interior angles. 65 + 55 = 120.

In review, A, B, and D are correct.

I hope this helps you =)

4 0

3 years ago

Read 2 more answers