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Dennis_Churaev [7]
3 years ago
6

If the area of a right angle triangle with base 3 cm is 6cm find the height​

Mathematics
1 answer:
gogolik [260]3 years ago
4 0

Answer:

A=1/2 Bh

6cm^2=1/2.3cm.h

12cm^2=3cmh

H=4cm

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In ΔABC, if m∠CAD = 29°, the m∠DAB is......
daser333 [38]

Answer:125

Step-by-step explanation:

6 0
3 years ago
What is the difference when the sum of 15 and 10 is subtracted from the product of 15 and 10.
Eva8 [605]

Answer:

125

Step-by-step explanation:

15+10=25

15x10=150

150-25=125

6 0
3 years ago
Read 2 more answers
A lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the do
wariber [46]

Answer:

There is not enough evidence to support the claim that the doors are either too long or too short.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 2058.0 millimeters

Sample mean, \bar{x} = 2047.0 millimeters

Sample size, n = 17

Alpha, α = 0.10

Sample standard deviation, s = 27

First, we design the null and the alternate hypothesis

H_{0}: \mu = 2058.0\text{ millimeter}\\H_A: \mu \neq 2058.0\text{ millimeter}

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{ 2047.0 - 2058.0}{\frac{27}{\sqrt{17}} } = -1.6798

Now, t_{critical} \text{ at 0.10 level of significance, 16 degree of freedom } = \pm 1.7396

Since, the calculated t statistic lies in the acceptance region, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to support the claim that the doors are either too long or too short.

5 0
4 years ago
Pls answer this I’m just gonna keep typing because I need to get to 20 pls help
DanielleElmas [232]

Answer:

its the second one hope this helps

Step-by-step explanation:

7 0
3 years ago
Please answer the question from the attachment.
spin [16.1K]

Hi,

7x²-2x-8=0, so a = 7, b= - 2, and c = -8.

You just need to apply the formula for square equations:

x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot 7\cdot (-8)}}{2\cdot 7}=\dfrac{2\pm\sqrt{228}}{14}=\dfrac{2\pm\sqrt{4\cdot 57}}{14}=\\\\=\dfrac{1\pm\sqrt{57}}7\Rightarrow x_1=\dfrac{1-\sqrt{57}}7,\;and\;x_2=\dfrac{1+\sqrt{57}}7.

Green eyes.

7 0
3 years ago
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