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bearhunter [10]
3 years ago
14

The value of 8 in 800 is how many times as large as the value of 8 in 80

Mathematics
2 answers:
MAVERICK [17]3 years ago
6 0
It is 10 times larger
Romashka [77]3 years ago
5 0
The value of 8 in 800 is 10 x as large as the value of 8 in 80
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Solve GHI. Round the answer to the nearest hundredth.
Sedaia [141]

Answer:

Part 1) HI=15\ units

Part 2)

Part 3) < I=28.07\°

The answer is the option A

Step-by-step explanation:

Part 1) Find the measure of side HI

Applying the Pythagoras Theorem

GI^{2} =GH^{2}+HI^{2}

substitute the values and solve for HI

17^{2} =8^{2}+HI^{2}

HI^{2}=17^{2}-8^{2}

HI^{2}=225

HI=15\ units

Part 2) Find the measure of angle G

In the right triangle GHI

cos(G)=\frac{GH}{GI}

substitute the values

cos(G)=\frac{8}{17}

G=arccos(\frac{8}{17})=61.93\°

Part 3) Find the measure of angle I

Remember that

The sum of the interior angles of a triangle must be equal to 180 degrees

so

substitute the values

61.93\°+90\°+< I=180\°

< I=180\°-(61.93\°+90\°)=28.07\°

7 0
3 years ago
Simplify the following expression.<br> -16 - 8/2 + 25 ÷ 5 + 1<br> -18<br> 6<br> 10<br> -6
Andrews [41]
The answer is -6. Thought there is a chance I am wrong. Need an explanation?
6 0
3 years ago
0.9 (13) - 1.7 (3) =
marissa [1.9K]

Answer:

6.6

Step-by-step explanation:

8 0
3 years ago
Distribute: -3(4a-5b+c)​
kvv77 [185]

Answer:

I got it.

Step-by-step explanation:

-12a - (-15b) + (-3c)

4 0
3 years ago
12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
Gnesinka [82]

\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

3 0
2 years ago
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