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2 years ago

Find the measure of x. 8.3 340 х x = [?] Round to the nearest tenth.

Mathematics

1 answer:

Tresset [83]2 years ago

3 0

Answer:

5.6 =x

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp side/ adj side

tan 34 = x / 8.3

8.3 tan 34 = x

5.59842 = x

To the nearest tenth

5.6 =x

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Sandy wants to make a punch recipe with 2 parts pineapple juice to 6 parts ginger ale if you wants to increase the recipe using

Blizzard [7]

Answer:

Step-by-step explanation:

If you go from 2 parts pineapple to 8, thats a 4x increase. Therefore 6x4=24 (6 from the Ginger)

4 0

3 years ago

Whats the area of this shape?

vodomira [7]

The area of the shape shown in the image which is in the shape of kite figure is 168 squared centimeters.

<h3>What is the area of the kite figure?</h3>

The area of the kite figure is half of the product of its diagonals. Area of kite figure can be find out using the following formula.

$A=\dfrac{pq}{2}$

Here, p and q are the diagonals of the kite.

The length of the first diagonal of the kite figure is,

p=6+15

p=21 cm

The length of the second diagonal of the kite figure is,

q=8+8

q=16 cm

Thus, the area of kite figure is:

$A=\dfrac{21\times16}{2}\\A=\dfrac{21\times16}{2}\\A=168\rm\; cm^2$

Thus, the area of the shape shown in the image which is in the shape of kite figure is 168 squared centimeters.

Learn more about the area of kite here;

brainly.com/question/16424656

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8 0

2 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago

Last year the dept of the river was 4.2 feet deep. THIS YEAR UT DROPPED 11% find the depth of the f the river this year

Romashka [77]

Answer:

new depth =

4.2/100 X 89 = 3.738 feet

4 0

3 years ago

Please explain how you found the answer.

otez555 [7]

35 divided by 5 = 7
7 x 2 = 14
So c) 14

And I have no idea about the second one sorry :/

8 0

3 years ago

Read 2 more answers