A 7/16 inch long bolt is used in a machine what is the length of the bolt written as a decimal

Mathematics

2 answers:

erik [133]3 years ago

6 0

The answer is 0.4375

stellarik [79]3 years ago

4 0

The answer is 0.4375. What you do is you divide the top number (numerator) by the bottom number (denominator) when you want to turn a fraction into a decimal. So 7 divided by 16 is 0.4375.

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1. Mrs. Verner's class has

Gemiola [76]

Answer:

46.7%

Step-by-step explanation:

Given:

Total number of students in Mrs. Verner's class = 15

Number of girls = 8

To find: percentage are boys

Solution:

Percentage of boys = ( Number of boys / Total number of students ) × 100

Number of boys = Total number of students - Number of girls = 15 - 8 = 7

So,

Percentage of boys = × 100 = 46.7%

4 0

2 years ago

If C = 2m2 + m and D = 2 – 6m + 2m², find an expression that equals<br> 2C - 2D in standard form.

lidiya [134]

C = 2m^2 + m
D = 2 - 6m + 2m^2

2C = 2(2m^2 + m) = 4m^2+2m
2D = 2(2-6m+2m^2) = 4-12m+4m^2

2C - 2D =
4m^2+2m-(4-12m+4m^2) =
4m^2+2m-4+12m-4m^2 =
0m^2 + 14m -4 =
14m - 4

3 0

3 years ago

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$