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2 years ago

If I got an 16/18 on my test,

Mathematics

1 answer:

aksik [14]2 years ago

5 0

Answer:

Step-by-step explanation:

Who knows

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Which of the following expression is equivalent to sin(5x) ?

RSB [31]

$Use:\\sin(x+y)=sinxcosy+sinycosx\\-------------------------\\sin5x=sin(3x+2x)=sin3xcos2x+sin2xcos3x\\\\Answer:e)$

5 0

2 years ago

Read 2 more answers

Need help with this question thanks :)

Fantom [35]

Answer:

Omar could invite 19 friends.

Step-by-step explanation:

the equation for this word problem would be

5x+30=125

subract 30 from both sides

5x=95

divide both sides by 5

x=19

8 0

3 years ago

Read 2 more answers

Value of (–3/8) × (+8/15). A. 11/15 B. 11/23 C. –1/5 D. –1/3

svet-max [94.6K]

ANSWER

The correct answer is C

EXPLANATION

We want to find the value of

$- \frac{3}{8} \times \frac{8}{15}$

This is the same as:

$- \frac{3}{8} \times \frac{8}{3 \times 5}$

We cancel out the common factors to get:

$- \frac{1}{1} \times \frac{1}{1 \times 5}$

We simplify to get:

$- \frac{1}{5}$

The correct answer is C

5 0

3 years ago

V^2 = u^2+ 2as u = 12 a = -3 s = 18 Work out a value of v.

Shkiper50 [21]

Answer:

use math-way

Step-by-step explanation:

6 0

3 years ago

Need help simplifying this

diamong [38]

The simplified answer is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$ .

Step-by-step explanation:

$$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term $(\frac{3x+y}{3x+y})$ and 2nd term by $(\frac{z+2x}{z+2x})$

= $\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}$

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= $\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}$

Now, the denominators are same, you can subtract it.

= $\frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

= $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

Thus the simplified solution is $\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}$

4 0

3 years ago