Ask question

Ask question

All categories

3 years ago

9

¿Q###N S#R# , S#R#, Q## D# N#CH# # V#C#S S#L# Y D# D## S# V#? adivinanza

2 answers:

Blababa [14]3 years ago

8 0

<span>who will be, it will be that night comes out and day goes...........</span>

bezimeni [28]3 years ago

4 0

Quien sera, sera que de noche sale y de dia se va?

You might be interested in

Hey guys I’m giving away free points today

pentagon [3]

Answer:

yayyyy!!!

Step-by-step explanation:

thanks that's very generous

3 0

3 years ago

Read 2 more answers

Suppose that the functions g and h are defined for all real numbers r as follows. gx) -4x +5 h (x) = 6x write the expressions fo

aleksandr82 [10.1K]

Answer: Our required values would be -10x+5, 2x+5 and -25.

Step-by-step explanation:

Since we have given that

g(x) = -4x+5

and

h(x) = 6x

We need to find (g-h)(x) and (g+h)(x).

So, (g-h)(x) is given by

$g(x)-h(x)\\\\=-4x+5-6x\\\\=-10x+5$

and (g+h)(x) is given by

$g(x)+h(x)\\\\=-4x+5+6x\\\\=2x+5$

and (g-h)(3) is given by

$-10(3)+5\\\\=-30+5\\\\=-25$

Hence, our required values would be -10x+5, 2x+5 and -25.

5 0

3 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

3 years ago

Work out:<br>6(10 - 23) + 20

barxatty [35]

Answer:

-58

Step-by-step explanation:

6 (10 - 23) + 20

=6 x (10 - 23) + 20

= 6 x (- 13) + 20

= -78 + 20

= -58

7 0

2 years ago

According to this diagram, what is sin 16°?<br> 74<br> 25<br> 16<br> 90<br><br> 24

velikii [3]

Answer:

7/25 sin= opposite/hypotenuse

Step-by-step explanation:

I'm just good like that.

4 0

3 years ago