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3 years ago

PLEASE ANY HELP I NEED HELP

Mathematics

1 answer:

miss Akunina [59]3 years ago

5 0

Answer:

(3×6) -1+4

18+3

21 is your answer ☺️. If I'm right so,

Please mark me as brainliest. thanks!!!

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How do I give an example of a positive and negative tempature that has a different of 5 degrees

love history [14]

Answer:

You could say +3 and -2 or +4 and -1

Step-by-step explanation:For example if you draw a number line

count 1 hump between each line. it equals 5.

|_______|_______|_______|_______|_______|

-2 -1 0 1 2 3

4 0

3 years ago

Sean says that to add a number to -100 and still have -100 is to add zero.candice says that she can add two numbers to -100 and

elena-s [515]

Both are right b/c like sean says you can add zero and itll still be -100 but candice is also right b/c because you can add any # to that -100 but you can also add a negative # to get it back to -100. for example: -100+23= -77 but if you also add a negative 23 it would be right back to -100.

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4 years ago

According to this equation, what is the value of x at t=0? express your answer in terms of the variables a, b, m, and k.

andriy [413]

Whats the equation?

5 0

4 years ago

Three right triangles are drawn in the coordinate plane,

SashulF [63]

Answer:

Triangle A:

- smaller leg = 3

- longer leg = 10

- hypotenuse = 10.4

Traingle B:

- smaller leg = 4

- longer leg = 13

- hypotenuse = 13.6

Triangle C:

- smaller leg = 5

- longer leg = 7

- hypotenuse = 8.6

Step-by-step explanation:

Let's find the distance of the legs for the 3 right triangles.

Triangle A

The smaller leg (SL) is from (2,5) to (2,8), so the length is given by the distance in the y-axis:

$SL_{A} = 8 - 5 = 3$

Hence, the lenght of the smaller leg is 3.

The longer leg (LL) is from (-8, 8) to (2, 8), so the length is given by the distance in the x-axis:

$LL_{A} = 2 - (-8) = 10$

Then, the length of the longer leg is 10.

Now, we can find the hypotenuse (H) by using Pitagoras:

$H_{A} = \sqrt{SL_{A}^{2} + LL_{A}^{2}} = \sqrt{3^{2} + 10^{2}} = 10.4$

Triangle B

The SL is from (3, -9) to (7, -9)

The length of the SL is:

$SL_{B} = 7 - 3 = 4$

The LL is from (7, -9) to (7, 4)

The length of the LL is:

$LL_{B} = 4 - (-9) = 13$

The hypotenuse is:

$H_{B} = \sqrt{4^{2} + 13^{2}} = 13.6$

Triangle C

The SL is from (-10, -6) to (-10, -1)

The length of the SL is:

$SL_{C} = -1 - (-6) = 5$

The LL is from (-10, -6) to (-3, -6)

The length of the LL is:

$LL_{C} = -3 - (-10) = 7$

The hypotenuse is:

$H_{C} = \sqrt{5^{2} + 7^{2}} = 8.6$

I hope it helps you!

5 0

4 years ago

Which of the following are identities? check all that apply.

Feliz [49]

Answer: (a), (b), (c), and (d)

Step-by-step explanation:

Check the options

$(a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x$

$(b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)$

$(c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x$

$(d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x$

Thus, all the identities are correct.

4 0

3 years ago