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3 years ago

(4.5)(5)(−2). simpilfy

Mathematics

1 answer:

Mariana [72]3 years ago

6 0

First do 4.5×5, which is 22.5
Then do 22.5×-2
Which is -45.
Hope this helps!

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Which graph represents the solution set of this inequality?<br> 11g-5< 49<br> Choose 1 answer:

N76 [4]

Answer:

Answer choice A is the correct answer

Step-by-step explanation:

** Before you start: If anything that is < or > you shouldn't fill in the plot.

If it has ≤ or ≥then you need to fill in the plot.**

Terminology:

(>) : Greater than (Don't fill in)

(<): Less than (Don't fill in)

(≤) : Less than or equal to (Fill in)

(≥) : Greater than or equal to (Fill in)

In this case, the answer choices C and D are wrong automatically.

Your Inequality:

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Subtract 5 both sides

11g ≤ 49-5

11g ≤ 44

Divide 44 by 11

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Find an answer choice that has a number greater than 4 but has a filled-in circle/plot.

Answer choice A is the correct answer

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2 years ago

Find the value of x ?

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3 years ago

Could you make proof of "a > b" please ?

Sergio039 [100]

Answer:

See explanation

Step-by-step explanation:

Consider the first triangle:

1. From the right triangle with acute angle $\theta_1:$

$\dfrac{c}{h}=\tan\theta_1\Rightarrow c=h\tan\theta_1$

2. From the right triangle with acute angle $\alpha:$

$\dfrac{c+a}{h}=\tan\alpha\Rightarrow c+a=h\tan\alpha$

Thus,

$h\tan\theta_1+a=h\tan\alpha\Rightarrow a=h\tan\alpha-h\tan\theta_1$

Consider the second triangle:

1. From the right triangle with acute angle $\theta_2:$

$\dfrac{d}{h}=\tan\theta_2\Rightarrow d=h\tan\theta_2$

2. From the right triangle with acute angle $\beta:$

$\dfrac{d+b}{h}=\tan\beta\Rightarrow b+d=h\tan\beta$

Thus,

$h\tan\theta_2+b=h\tan\beta\Rightarrow b=h\tan\beta-h\tan\theta_2$

Now, since $\alpha>\beta,$ we have $\tan\alpha>\tan\beta$ and $\sin\alpha>\sin\beta.$

$\dfrac{\sin\alpha}{\sin\theta_1}=\dfrac{\sin\beta}{\sin\theta_2}$

and

$\sin\alpha>\sin\beta,$

then

$\sin\theta_1>\sin\theta_2.$

This means $\theta_1>\theta_2$ and $\tan\theta_1>\tan \theta_2.$

Hence,

$h\tan\theta_1>h\tan \theta_2\\ \\-h\tan\theta_1$

Now,

$h\tan\beta$

and

$-h\tan\theta_1$

$h\tan\beta-h\tan\theta_1$

Note that this solution is true only for acute angles $\alpha,\ \beta,\ \theta_1,\ \theta_2$

6 0

3 years ago