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3 years ago

What is the length h of the right triangle, rounded to the nearest tenth?

Mathematics

2 answers:

gavmur [86]3 years ago

4 0

Sin(35)x14 =8.03007...
So answer is 8

Natasha_Volkova [10]3 years ago

4 0

Answer:

Step-by-step explanation:

sin35=opp/hyp

opp=sin35(14)=8.03 close to 8

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A null hypothesis is being tested. How does the confidence level affect the testing process?

xxTIMURxx [149]

If you have a high confidence level, the chance of rejecting the null hypothesis is rare. If you have a low confidence level, the chance of of rejecting the null hypothesis is nonexistent. If you have a low confidence level, the chance of of rejecting the null hypothesis is rare. If you have a high confidence level, the chance of of rejecting the null hypothesis is high.

5 0

3 years ago

Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find

densk [106]

Answer:

x=4

Step-by-step explanation:

Step 1:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

$AC^2 = AB^2 + BC^2$

step 2:-

The hypotenuse is longest side is AC = 17 units

(17)^2 = 4x^2 +9(X+1)^2

on simplification, we will use formula

$(a + b)^2 = a^2 +2ab+b^2$

289 = 4x^2 +9(x^2+2x+1)

$13x^2 +18x-280 = 0$

finding factors 70 X 52 = 3640

$13x^2 +70x-52x-280 = 0$

$13x^2 -52x+ 70x-280 = 0$

Taking common , we get

13x(x-4)+70(x-4)=0

x-4=0 and 13x+70=0

x=4 and $13x =-70$

x=4 and $x=\frac{-70}{13}$

we can not choose negative value so x value is 4

Final answer:- x = 4

verification:-

$AC^2 = AB^2 + BC^2$

289 = 4(4)^2+9(4+1)^2

289 = 64 +9(25)

289=289

6 0

3 years ago

Read 2 more answers

How to divide the area of 10X10 into 30 equal pieces

-BARSIC- [3]

30 divided by 10 and youll get it into each hope that helped

7 0

3 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

6 0

3 years ago

Solve the inequality and graph the solution a b c or d

Kruka [31]

Answer: I pretty sure it’s D or A but I’m going with A

Step-by-step explanation: I hope this helps you and I know I’m kinda late but I still hope this helps! Have a great day!

4 0

3 years ago