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Andrej [43]
1 year ago
6

I am in need of some help :(It looks so simple yet I don't get how to do it. Can anyone help explain?(picture of question attach

ed)source: Pearson/Savvas Realize

Mathematics
1 answer:
Firdavs [7]1 year ago
5 0

x = 14

Explanation:

height = x

base 1 = 98

base 2 = 2

Using altitude formula in geometric mean:

base 1/height = height/base 2

98/x = x/2

\begin{gathered} \frac{98}{x}=\frac{x}{2} \\ \text{cross multiply:} \\ 2(98)\text{ = x}\times x \end{gathered}\begin{gathered} 196=x^2 \\ \text{square root both sides:} \\ \sqrt[]{196}\text{ = }\sqrt[]{x^2} \\ x\text{ = }\sqrt[]{196} \\ x\text{ = 14} \end{gathered}

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vladimir2022 [97]

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10

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1/2×(-2-(-7))×(6-(-4))

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3 years ago
Evaluate the expression P(6,1)
klasskru [66]

Answer:

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Step-by-step explanation:

In the pic is how to do it

5 0
3 years ago
I need help asap someone please help me i dont understand this question so can someone help me
eduard

Answer:

we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

Step-by-step explanation:

Given the expression

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}

=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}

\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}

=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}

cancel the common factor: 2

=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}

cancel the common factor: p

=\frac{2p+1}{p^2\left(4p^2-1\right)}

=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}

cancel the common factor: 2p+1

=\frac{1}{p^2\left(2p-1\right)}

Expanding

=\frac{1}{2p^3-p^2}

Thus, we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

7 0
3 years ago
30 Between what two standard deviations of a normal distribution contain 68% of the data?
Lady bird [3.3K]

d

man i think its d or b maby

4 0
3 years ago
I need to find the zeros to these two problems and I have no idea how
Iteru [2.4K]
If you have a graphing calculator just put in the equation in 'y=' (not the i equation), and then go to 2nd trace and see where the y=0, those numbers under the x column are the zeros. For the first one, the zeros are: -1, .5, and 2.8. For the second question the zeros are: -3 and about 1.9. The zeros with a decimal are estimations.
7 0
3 years ago
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