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3 years ago

14

ricardo have two snacks.One snack has the mass of 167 grams,and the other has 243 grams what is the difference in the mass of sn

acks

1 answer:

Fed [463]3 years ago

3 0

243 - 167 =76
The difference is 76g.
You just need to find the difference between both snacks

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A basketball hoop is purchased for use outside. It is on a moveable

Umnica [9.8K]

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49 + 32 = 81 81 divided by 9 is 9 in

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2 years ago

Find the x -intercept and y -intercept of the line. 2x-6y=18

lapo4ka [179]

The x intercept is 9
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3 years ago

A rectangle has an area of 20 ft.² in a similar rectangle has an area of 180 ft.² what is the ratio of areas of the similar

vovikov84 [41]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{cccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array}\\\\
-----------------------------\\\\$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s^2}{s^2}=\cfrac{20}{180}\implies \left( \cfrac{s}{s} \right)^2=\cfrac{20}{180}\implies \cfrac{s}{s}=\sqrt{\cfrac{20}{180}}
\\\\\\
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3 years ago

HELP PLZZ will give brainliest <3

melisa1 [442]

Answer:

0

1

Step-by-step explanation:

First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

$\dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B}$

$\dfrac{1}{0.5} = \dfrac{4}{\sin B}$

$\sin B = 2$

The sine function can never equal 2, so there is no triangle in this case.

Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

$\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}$

$\sin C = \dfrac{8.9\sin 63^\circ}{10}$

$C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}$

$C \approx 52.5^\circ$

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.

Answer: one triangle

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3 years ago

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nasty-shy [4]

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