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3 years ago

10

(6²+2²)÷4−10 Can someone HElp?

2 answers:

marissa [1.9K]3 years ago

8 0

Answer:

The answer is -20/3

Step-by-step explanation:

(6² + 2²) ÷ 4 - 10

36 + 4 ÷ (-6)

40/(-6) = -20/3

Thus, The answer is -20/3

<u>-</u><u>TheUnknownScientist</u><u> 72</u>

Akimi4 [234]3 years ago

4 0

Answer:

0

Step-by-step explanation:

6^2 + 2^2 = 36 + 4 = 40 Evaluate what is in the brackets.

40/4 = 10 Divide by 4

10 - 10 = 0 Subtract 10

The answer is zero.

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Draw counters to make 10 using number 7 and 8 .

Eva8 [605]

You could take @ birthday candle (7&8) and warm the wax to draw a 10 on your paper or if u need to write 7 and 8 in the shape of a counter, u could either:

1) draw 7 and 8 on your paper and circle the parts that look like 10 7 could be a slanted 1 and 8 could be your 0

2) you could draw parts of 7 and 8 to make it look like 10

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4 years ago

F(x) = 3x2 - 7 and g(x) = x3 + 1, evaluate f(g(2)). NEED HELP PLEASEE??

Triss [41]

F(g(2)) = 3(x^3 + 1)^2 - 7 = 3(x^6 + 2x^3 + 1) - 7 = 3x^6 + 6x^3 + 3 - 7 = 3x^6 + 6x^3 - 4

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3 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

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If a 36-inch yardstick casts a 21-foot shadow, how tall is a building whose shadow is 168 feet?

MAVERICK [17]

The building would have to be 24 feet tall

36 inch is 3 ft
21/3=7
168/7=24

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3 years ago

SOS<br>Find X in the 45 -45 -90 triangle.

Alja [10]

Answer:

1.697

Step-by-step explanation:

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3 years ago