Ask question

Ask question

All categories

3 years ago

13

Picture attached please answer quickly

2 answers:

Fynjy0 [20]3 years ago

5 0

i feel so bad for you, i don't even know that.

DedPeter [7]3 years ago

4 0

Answer:

The last option!!!!

You might be interested in

Please help me with question 24 ‍♀️

love history [14]

for answer 2 is A=1/32

for answer 1 is=A=4.8

hope that helps you

3 0

3 years ago

For 9-20 tell whether the first number is a multiple of the second number

jeyben [28]

Answer:

21,7 - yes

28,3 - no

17,3 - no

20,4 - yes

55,5 - yes

15,5 - yes

26,4 - no

32,8 - yes

48,7 no

60,2 - yes

79,4 - no

81,3 - yes

Step-by-step explanation:

hope that helps

7 0

2 years ago

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

$\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|$

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with

$\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4$

3 0

3 years ago

Solve x+5<-5 for x<br> please help

yawa3891 [41]

Answer:

x<-25

Step-by-step explanation:

5 0

3 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago