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3 years ago

How do you use pythagorean theorem finding c on a right triangle with one side 10 and the other 5?

2 answers:

8 0

The pythagorean theorem is C^2=a^2+b^2.(^2=squared) Where C is equal to the longest side of the of the triangle or the hypotnos. So if you plug 10 into a and 5 into b then your formula is C^2=10^2+5^2 simplifying to C^2=100+25 simplifying again C^2=125. Now you don't want C^2 you just want c so you have to take the square root of both sides. sqrt(c^2)=sqrt(125) simplifying to c=11.180 if you round to the 3rd decimal place. Final answer c=<span>11.180.</span>

Vinvika [58]3 years ago

7 0

Just remember: A^2 + B^2 = C^2 (^2 = squared or raised to the second). Also. A and B are either of the legs. C is 100% of the time the hypotenuse! The hypotenuse is the longest side of a triangle and will always be located directly across from the right angle symbol (the 90 degrees marker). If there is no marker, then the triangle is not a right angle triangle, and the Pythagorean Theorem cannot be used.

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Answer:

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The radius of the circular track was 157.64 feet find the circumference of the track

DerKrebs [107]

Before we solve for anything, let's remember what our formula is.

The circumference of a circle is 2PiR (2 x Pi x R).

Our radius is 157.64.
Pi is 3.14 (estimated).

Let's plug our numbers into the formula.

Circumference = 2 x 3.14 x 157.64.
2 x 3.14 = 6.28
6.28 x 157.64 = 989.979 (990 if estimated).

Your circumference is:
989.979 ft^2.

I hope this helps!

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