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2 years ago

5.

Mathematics

2 answers:

tresset_1 [31]2 years ago

8 0

A and D because it is basically saying 3:36 so when you simplify it comes down to 1:12

Marizza181 [45]2 years ago

6 0

It’s lowkey A i think

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Question 25 of 25

frez [133]

Answer:

It is actually 2 and -6.

Step-by-step explanation:

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3 years ago

Help with both plz I’ll mark brainliest

LekaFEV [45]

Answer:

(1,6) is not a solution and I think the second one is: c I think

Step-by-step explanation:

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2 years ago

What is the value of x in the eqation 5x+3=4x

Reika [66]

Answer :x=-3

Step-by-step explanation:Solve for x by simplifying bothh sides of the equation then isoolating the variable

7 0

2 years ago

Read 2 more answers

Is the statement true or false? If false, give a counterexample.

scoundrel [369]

Answer:

D. false; if a = 1, b = 2, and c = 3, then 1(2 + 3) ≠ 1(2) + 2(3)

Step-by-step explanation:

A is false, because the a is being distributed/ being multiplied to all terms inside the parenthesis, and not the term b.

B is also false. There is no indicated negative signs.

C is also false because 1(1 + 1) is EQUAL to 1(1) + 1(1)

D is true.

6 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

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3 years ago