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1 year ago

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Mark is making a decoration for a rally, using a string of triangular strips. Each strip is an isosceles triangle when flattened

out.
( by the way the picture is a triangle that is 70 degrees)

What is the measure of the third angle?

1 answer:

jeyben [28]1 year ago

5 0

Solution:

Consider the Given Isosceles Triangle

Considering the Possibilities

Case 1. When two equal angles are of 70°

Let the third angle be x.

Keeping in mind , that sum of Interior angles of Triangle is 180°.

70° + 70° + x= 180°

140° +x= 180°

x= 180°- 140°

x= 40°

Case 2:

When an angle measures 70°, and two equal angles measure x°.

Keeping the same property of triangle in mind, that is sum of interior angles of triangle is 180°.

70° + x° + x° = 180°

⇒ 70° + 2 x° = 180°

⇒ 2 x° = 180° - 70°

⇒ 2 x° = 110°

Dividing both sides by 2, we get

x= 55°

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The table shows the side length and approximate area of an octagonal stop sign.

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2 years ago

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Ann is grouping 38 rocks. She can put them into groups of 10 rocks or as single rocks. What are the different ways Ann can group

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Answer: There will be two different ways of doing so.

Step-by-step explanation:

Since we have given that

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<u>Case I:</u>

if we group into 10 rocks

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and Fourth set remains with 8 rocks only.

<u>Case II:</u>

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2 years ago

Suppose that Randy is an analyst for the bicyling industry and wants to estimate the asking price of used entry-level road bikes

padilas [110]

Answer: The lower limit of this confidence interval = $576.41

The upper limit of this confidence interval = $831.09

Step-by-step explanation:

Let $\mu$ be the population mean price of a used road bike.

As per given have,

n=11

Degree of freedom = 10

$\overline{x}=\$703.75$

s= $189.56

T-critical value for 95% confidence :

$t_{(df, \alpha/2)}=t_{(10,0.025)}=2.228$

Now, 95% confidence interval for μ, the mean price of a used road bike. will be :

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

$\$703.75\pm (2.228)\dfrac{189.56}{\sqrt{11}}$

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$(\$703.75-\$127.34,\ \$703.75+\$127.34)$

$(\$576.41,\ \$831.09)$

Thus , the lower limit of this confidence interval = $576.41

The upper limit of this confidence interval = $831.09

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1 year ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as

lubasha [3.4K]

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

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1 year ago