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3 years ago

.please help me with this I will crown you brainlest

Mathematics

2 answers:

Brut [27]3 years ago

8 0

I think it is either A or C
but more sure about A

Marizza181 [45]3 years ago

4 0

You can prove it by the SAS postulate which should be the first option

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Suppose you can factor x^2 + bx + c as (x + p)(x + q). If c > 0, what could be possible values of p and q?

ad-work [718]

p could be 5 and q could be 8 this is just my guess but i think its right

3 0

2 years ago

Simplify this ratio completely 24/16

Bas_tet [7]

Answer:

3/2 or 1.5

Step-by-step explanation:

24/16 = 12/8

12/8 = 6/4

6/4 = 3/2

So, if you want fraction it's 3/2 and decimal is 1.5

5 0

2 years ago

The difference of twice a number g and 10 is 24

ANTONII [103]

The number g is 7
24-10 = 14
14/2=6

8 0

3 years ago

Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su

Alja [10]

Answer:

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

Step by step explanation:

$\frac{ {(4 {n}^{4} {q}^{5})}^{2} {(8 {n}^{4} q)}^{-2} }{ {(- 3 {nq}^{9})}^{ - 1} {(4 {n}^{3} {q}^{9}) }^{3} }$

first we will change the terms with negative superscrips to the other side of the fraction

$\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }$

then we will distribute the superscripts

$\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2} {q}^{2} }$

$\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8} {q}^{2} }$

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

${4}^{2 - 3} {q}^{10 + 9 - 2 - 27} {n}^{8 + 1 - 8 - 9} {8}^{ - 2} { (- 3)}^{1}$

${4}^{ - 1} {q}^{ - 10} {n}^{ - 8} {8}^{ - 2} { (- 3)}^{1}$

then we will change again the terms with negative superscrips to the other side of the fraction

$\frac{ - 3}{ 4 \times {8}^{2} {q}^{10} {n}^{8} }$

$\frac{ - 3}{ 256 {q}^{10} {n}^{8} }$

4 0

2 years ago

Easy make sure its detailed 15 points

goblinko [34]

9514 1404 393

Answer:

no square roots: -1000, -8
one square root: 0
two square roots: 8, 64, 1000
no cube roots: <none>
one cube root: -1000, -8, 0, 8, 64, 1000
two cube roots: <none>

Step-by-step explanation:

The attached graph shows the square root relation (red) and the cube root function (blue). The function values are shown for x=0 and x=±8.

You can see that there are 2 square roots for positive numbers, one square root for 0, and 0 square roots for negative numbers. There is exactly 1 cube root for any number.

no square roots: -1000, -8
one square root: 0
two square roots: 8, 64, 1000
no cube roots: <none>
one cube root: -1000, -8, 0, 8, 64, 1000
two cube roots: <none>

_____

Additional comment

We call the square root curve a "relation" because it is not a function. A relation that is a function will have only one y-value for each x-value. For positive x-values, there are two square roots.

4 0

2 years ago