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Semenov [28]
3 years ago
15

How do I solve

Mathematics
1 answer:
Maslowich3 years ago
4 0

Use:\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\\\\sqrt[n]{a}=a^\frac{1}{n}\\------------------------\\\\\left(16n^4\right)^{\frac{5}{4}}=16^\frac{5}{4}\left(n^4\right)^\frac{5}{4}=16^{1\frac{1}{4}}n^{4\cdot\frac{5}{4}}=16^{1+\frac{1}{4}}n^5=16^1\cdot16^\frac{1}{4}\cdot n^5\\\\=16\cdot\sqrt[4]{16}\cdot n^5=16\cdot2\cdot n^5=\boxed{32n^5}\\\\\sqrt[4]{16}=2\ because\ 2^4=16\\\\\text{Answer}\ \boxed{\left(16n^4\right)^\frac{5}{4}=32n^5}

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STatiana [176]

Answer:

x=-8

Step-by-step explanation:

Subtract 2 from both sides

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2 years ago
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Explanation

Answer:

The carpenter can build the triangular frame by cutting the 12 foot piece by 2foot

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1 year ago
What is the product 3x5(2x2+4x+1)
lesya692 [45]

Answer:

6x^7+12x^6+3x^5

Step-by-step explanation:

The given expression is:

3x^5(2x^2+4x+1)

We expand the parenthesis using the distributive property to obtain;

3x^5(2x^2)+3x^5(4x)+3x^5(1)

Recall that;

a^m\imes a^n=a^{m+n}

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6x^7+12x^6+3x^5

3 0
3 years ago
What is the area of the trapezoid?
Maslowich
<h3><u>Given Information :</u></h3>

  • Length of parallel sides = 60 ft and 40 ft
  • Height of the trapezoid = 30 ft

<h3><u>To calculate :</u></h3>

  • Area of the trapezoid.

<h3><u>Calculation :</u></h3>

As we know that,

\bigstar \: \boxed{\sf {Area_{(Trapezium)} = \dfrac{1}{2} \times ( a + b) \times h}} \\

  • a and b are length of parallel sides.
  • h denotes height.

<em>S</em><em>u</em><em>b</em><em>s</em><em>t</em><em>i</em><em>t</em><em>u</em><em>t</em><em>i</em><em>n</em><em>g</em><em> </em><em>valu</em><em>es</em><em>,</em><em> </em><em>we</em><em> </em><em>get</em><em> </em>:

\longmapsto Area = \sf \dfrac{1}{2} × ( 60 + 40 ) × 30 ft\sf ^2

\longmapsto Area = \sf \dfrac{1}{2} × 100 × 30 ft\sf ^2

\longmapsto Area = 1 × 100 × 15 ft\sf ^2

\longmapsto Area = 100 × 15 ft\sf ^2

\longmapsto <u>Area = 1500 ft</u>\sf ^2

Therefore,

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6 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

8 0
3 years ago
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