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Maru [420]
1 year ago
5

11. Find the value of x. x=______

Mathematics
1 answer:
Tems11 [23]1 year ago
4 0

Answer: x = 3.5

Step-by-Step Solution:

Let us first label the figure.

Let the Triangle be ABC with a line DE || BC.

Now, in ∆ABC,

DE || BC (given)

=> AD/DB = AE/EC (by B.P.T)

Substituting the given values,

AD/DB = AE/EC

2/4 = x/7

1/2 = x/7

2x = 7

x = 7/2

=> x = 3.5

Therefore, x = 3.5

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What is 8.05x10^-5 in standard notation
Marina CMI [18]

Answer: 0.0000805

Step-by-step explanation:

You move the decimal 5 places to the left of 8.05.

6 0
3 years ago
Helppppppppppp meeeeeeeeeeeeeeeee give bralienst
Eduardwww [97]

Answer:

Point C

Step-by-step explanation:

Point c is the only point on the number line which is in between 2 and 3.

<em>Thus,</em>

<em>point c is the answer.</em>

<em>Hope this helps :)</em>

4 0
3 years ago
Please help and explain
mr_godi [17]
All but -1 <x <1
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4 0
3 years ago
A Chemist needs to mix a 20% acid solution with a 50% acid solution to obtain 15 liters of a 34% acid solution. How many liters
Nuetrik [128]

8 litres (amount of 20% solution needed) and 7 litres for (amount of 50% solution needed)

<u>Step-by-step explanation:</u>

Let consider ‘x’ for 20% acid solution and (15 – x) for 50% acid solution. And so, the equation would be as below,

                         20% in x + 50% in (15 – x) = 15 litres of 34%

Convert percentage values, we get

                              0.20(x) + 0.50 (15 – x) = 15 (0.34)

                              0.20 x + 7.5 – 0.50 x = 5.1

                              -0.3 x + 7.5 = 5.1

                              0.3 x = 7.5 – 5.1

                              0.3 x = 2.4

x = \frac{2.4}{0.3} = 8 litres (amount of 20 \% solution needed)

Apply ‘x = 8’ value in (15 – x) we get,

                         15 – 8 = 7 litres

The value of 7 litres for (amount of 50% solution needed)

8 0
3 years ago
Given: ΔABC; b= 10; c = 14, and ∠A = 54°. Find the length of side a to the nearest whole number.
alexandr402 [8]
A² = b² + c² - 2bc cosA
a² = 10² + 14² - 2*10*14 cos54
a² = 100 + 196 - 280 * cos54
a = \sqrt{100 + 196 - 280 * cos54}
a = 11.46

5 0
4 years ago
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