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2 years ago

Question 1

Mathematics

1 answer:

Over [174]2 years ago

8 0

Step-by-step explanation:

82 + 16 = 98

Expression A: 122 - 82 = 40

Expression B: 2(62) + 23 = 147

Expression C: 102 - 32 = 70

None of them has the same value as 82 + 16.

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3 3/4+ (-4 1/3) Simplify

IgorC [24]

Answer:

-7/12 in decimal form it would be -0.583

Step-by-step explanation:

convert the mixed number to an improper fraction

(when there's a + in front of an expression in parentheses the expression remains the same)

15/4+(-4 1/3)

convert the mixed number to an improper fraction

15/4-13/3

subtract the fractions to get -7/12

4 0

3 years ago

If f(x) = x, then f(64) is equal to a. 1/4 b. -8 c. -4 d. 4

blsea [12.9K]

B hope it helps have a good day

6 0

3 years ago

Express sin A,cos A and tan A as ratios

OverLord2011 [107]

Answer:

Part A) $sin(A)=\frac{2\sqrt{42}}{23}$

Part B) $cos(A)=\frac{19}{23}$

Part C) $tan(A)=\frac{2\sqrt{42}}{19}$

Step-by-step explanation:

Part A) we know that

In the right triangle ABC of the figure the sine of angle A is equal to divide the opposite side angle A by the hypotenuse

$sin(A)=\frac{BC}{AB}$

substitute the values

$sin(A)=\frac{2\sqrt{42}}{23}$

Part B) we know that

In the right triangle ABC of the figure the cosine of angle A is equal to divide the adjacent side angle A by the hypotenuse

$cos(A)=\frac{AC}{AB}$

substitute the values

$cos(A)=\frac{19}{23}$

Part C) we know that

In the right triangle ABC of the figure the tangent of angle A is equal to divide the opposite side angle A by the adjacent side angle A

$tan(A)=\frac{BC}{AC}$

substitute the values

$tan(A)=\frac{2\sqrt{42}}{19}$

8 0

3 years ago

Which of the following Triangle cases cannot be solved, using either the law of cosines or law of sines

hodyreva [135]

The answer is D. You cannot use aaa

8 0

3 years ago

Read 2 more answers

When the solutions are going to be complex, which methods can be used to solve quadratic equations?

yKpoI14uk [10]

You could complete the square to state the vertex.
You could use the quadratic equation to find the roots (which are complex).

Try an example that will require both.

y = x^2 + 2x + 5

Step One
Get the graph. That's included below.

Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)

y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4

The vertex is at (-1,4)

Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

a = 1
b = 2
c = 5

$\text{x = }\dfrac{ -2 \pm \sqrt{2^{2} - 4*1*5 } }{2}$
$\text{x = }\dfrac{ -2 \pm \sqrt{4 - 20 } }{2}$
$\text{x = }\dfrac{ -2 \pm \sqrt{-16 } }{2}$
$\text{x = }\dfrac{ -2 \pm 4i }{2}$
x = -1 +/- 2i

x1 = -1 + 2i
x2 = -1 - 2i And we are done.

7 0

3 years ago