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pantera1 [17]
3 years ago
5

Need answer ASAP. Worth 25 points

Mathematics
1 answer:
kompoz [17]3 years ago
5 0

I think the answer is B and C

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1. Kelly has a rock garden with a length of 6 feet. She constructs a scale model of the rock garden using the scale 1 inch:2 fee
alexandr1967 [171]

Answer:

<em>(a) 3 inches </em>

<em>(b) Her scale model drawing WILL fit on the piece of paper 8.5 by 7 inches</em>

Step-by-step explanation:

<u>Scaling </u>

We use scales to represent realities in reduced spaces. In the case of lengths or surfaces, the scale allows us to draw the shapes in normal sheets of paper.  

(a)

Kelly's rock garden has a length of 6 feet. She uses a 1 inch:2 feet scale. this means that the model length is

6 feet*1 inch/2 feet = 3 inches

(b)

The width of the model is 5 inches, keeping the same scale it means the real length is

5 inches * 2 feet / 1 inch = 10 feet

The model will have 3 inches by 5 inches, it perfectly fits on any regular piece of paper

6 0
3 years ago
Which angle has a sine of -1/2 and a cosine of -√3/2
Sergeeva-Olga [200]
\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)

\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}

\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}

x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}
8 0
3 years ago
Please help me on the first question
tino4ka555 [31]

Answer:

x=5

Step-by-step explanation:

If you add the like terms, 180 would equal 35x + 5. If you subtract 5 on each side, you would get 35x = 175. If you divide each ide by 35 to isolate x, x would be 5.

6 0
3 years ago
Read 2 more answers
From least to greatest 8/100 3/5 7/10
stich3 [128]
8/100, 3/5, 7/10

Explanation:
8/100 = 0.08
3/5 = 0.6
7/10 = 0.7
6 0
3 years ago
Read 2 more answers
Guys Stan lee is dead..... I need ur help
kipiarov [429]
Sure how can I help?
6 0
3 years ago
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