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2 years ago

How to solve this problem

Mathematics

2 answers:

Klio2033 [76]2 years ago

4 0

Answer:

there is no picture or anything

insens350 [35]2 years ago

3 0

What’s da problem that u have?

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What is p?<br><br> 7 1/5+p+4 9/10+3 3/20=19 1/10

Yakvenalex [24]

ANSWER:
$7 \frac{1}{5}+p+4 \frac{9}{10}+3 \frac{3}{20} =19 \frac{1}{10} \\ 7 \frac{4}{20}+p+4 \frac{18}{20} + 3 \frac{3}{20}=19 \frac{2}{20} \\ p=19 \frac{2}{20}- (7 \frac{4}{20}+4 \frac{18}{20} + 3 \frac{3}{20}) \\ p=19 \frac{2}{20}-15 \frac{5}{20} \\ p=3\frac{17}{20}$

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3 years ago

if triangle ABC is an isosceles triangle and triangle DBE is an equilateral triangle, find each missing measure

BabaBlast [244]

Picture?? This question is incomplete just get a picture then I can answer it

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3 years ago

Simplify (-2)(-3)^2-2(2 - 5) <br><br>39 -12 -24

daser333 [38]

Answer:

-12

Step-by-step explanation:

(-2)(-3)^2-2(2-5). Original

-2(9)-2(2-5) Distribute Exponents

-18-4+10 Distribute Parentheses

-22+10 Do subtraction

-12 Do Addition

4 0

3 years ago

Read 2 more answers

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago

Can any expert help me plzzzzzzzz..........

natali 33 [55]

Answer:

i=f

ii= 3-1/20

iii=3.43234

iv=a

v=12

Step-by-step explanation:

6 0

3 years ago