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2 years ago

What number is 76% of 97

Mathematics

2 answers:

IrinaVladis [17]2 years ago

7 0

Answer:

73.72

Step-by-step explanation:

76x96/100=73.72

alexandr1967 [171]2 years ago

6 0

76%*97
=73.72

I hope this helps.

Comment if you need further explanation.

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If a gardner plants beans in 1/8 of a plot and tomatoes in 1/6 of a plot, what fraction of the garden is left for other crops

JulsSmile [24]

You simply have to sum the two fraction together to get the total amount of occupied plot. To do so, multiply both numerator and denominator of each fration by the same number, so that they end up having the same denominator.

You can multiply both numerator and denominator of 1/8 by 3 to get 3/24. Similarly, you can multiply both numerator and denominator of 1/6 by 4 to get 4/24.

Now we can easily sum the two fractions together:

$\dfrac{1}{8}+\dfrac{1}{6}=\dfrac{3}{24}+\dfrac{4}{24} = \dfrac{7}{24}$

So, if 7 parts out of 24 are occupied, it means that there are $24-7=17$ parts left for other crops. So, the answer is 17/24.

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2 years ago

In the triangle shown we can find the angle θ as follows calculator

Xelga [282]

Given a right triangle with hypothenus of measure 34, the side opposite the angle θ of measure 30, and the side adjacent the angle theta of measure 16.

$\sin\theta= \frac{opposite}{hypothenuse} = \frac{30}{34} = \frac{15}{17} \\ \\ \Rightarrow \theta=\sin^{-1}\left( \frac{15}{17} \right) \\ \\ \\ \cos\theta= \frac{adjacent}{hypothenuse} = \frac{16}{34} = \frac{8}{17} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{8}{17} \right) \\ \\ \\ \tan\theta= \frac{opposite}{adjacent} = \frac{30}{16} = \frac{15}{8} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{15}{8} \right)$

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3 years ago

Suppose you have a die that has 20 sides numbered 1 to 20. What is the probability the

Digiron [165]

Answer:

$\frac{1}{10}$

Step-by-step explanation:

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2 years ago

This is a dumb question but what is 68 time 100

Whitepunk [10]

68 x 100 = 6,800

Hope this helps

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2 years ago

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How do i prove 2sec^2x-2sec^2xsin^2-sin^2x-cos^2x=1

kramer

$2\sec^2x-2\sec^2x\sin^2x-\sin^2x-\cos^2x=2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)$

Since $\sin^2x+\cos^2x=1$ , you have

$2\sec^2x(1-\sin^2x)-(\sin^2x+\cos^2x)=2\sec^2x\cos^2x-1$

and since $\sec x=\dfrac1{\cos x}$ , you end up with $2-1=1$ .

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2 years ago