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3 years ago

X − 5 = 17 A) x = 22 B) x = 12 C) x = 3.4 D) x = -12

Mathematics

1 answer:

Maksim231197 [3]3 years ago

8 0

A is the correct answer coz if you substitute 22 in the place of X-5 the answer will be 17

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1 = 18, W= 10. h = 2

leonid [27]

Answer:

Step-by-step explanation:

yes ,great job lxwxh

6 0

3 years ago

Let AB = 24, AC = 10 and BC = 26.

nata0808 [166]

Answer:

a) 24

b) 10

c) 12/13

d) 5/13

e) 12/5

Step-by-step explanation:

a) We can see that the leg opposite <C is AB, and we are given AB = 24

b) We can see the leg adjacent to <C is AC, and we are given that AC = 10

c) The trig function sine is equal to

$\frac{opposite}{hypotenuse}$

The opposite, AB, is 24, and the hypotenuse, BC, is 26. We can plug those numbers in:

$sin(c) = \frac{24}{26} = \frac{12}{13}$

d)The trig function cosine is equal to

$\frac{adjacent}{hypotenuse}$

The adjacent, AC, is 10, and the hypotenuse, BC, is 26. We can plug those numbers in:

$cos(c) = \frac{10}{26} = \frac{5}{13}$

d)The trig function tangent is equal to

$\frac{opposite}{adjacent}$

The opposite, AB, is 24, and the adjacent, AC, is 10. We can plug those numbers in:

$tan(c) = \frac{24}{10} = \frac{12}{5}$

4 0

3 years ago

What is 48% of 64 please give right answers only

Charra [1.4K]

Answer:

30.72

Step-by-step explanation:

7 0

3 years ago

Please show your work!!!!

viva [34]

Answer:

4√34

Step-by-step explanation:

Let the unknown side be y

y^2 = 20^2 + 12^2

y^2 = 400 + 144

y^2 = 544

Take the square root of both side

y = √544

y = √(16x34)

y = √16 x √34

y = 4√34

6 0

3 years ago

I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have

$i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079$

3 0

3 years ago