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3 years ago

Which quotient is greater than 2? 50 ÷ 28 45 ÷ 23 38 ÷ 16 28 ÷ 15

Mathematics

2 answers:

mezya [45]3 years ago

8 0

38/16 is the answer

DerKrebs [107]3 years ago

4 0

Answer: $Hello, 38 divided by 16 equals a quotient greater than 2$

Step-by-step explanation:

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Planes X and Y are perpendicular. Points A, E, F, and G are points only in plane X. Points R and S are points in both planes X a

marusya05 [52]

Answer:

Planes EA and FG

Step-by-step explanation:

3 0

3 years ago

How many solutions does the following equation have? ln(x2 + 4x − 5) = 0

lisov135 [29]

Alrighty
remember
$log_a(b)=c$ means $a^c=b$
and
$ln(x)=log_e(x)$
and
$x^0=1$ for all real values of x
so

$ln(x^2+4x-5)=0$ means
$log_e(x^2+4x-5)=0$ means
$e^0=x^2+4x-5$ which simplifies to
1=x²+4x-5
minus 1 both sides
0=x²+4x-6
use quadratic formula or complete the square
$x=-2+\sqrt{10}$ and $x=-2-\sqrt{10}$
2 solutions

7 0

3 years ago

HELP TIMED BRAINLY POINTS

kozerog [31]

Answer:

dude you need to show the whole answer

Step-by-step explanation:

sorry I couldn't help

6 0

3 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago

Tina wrote the equations 3x - y = 9 and 4x + y + 5, What can Tina conclude about the solution to this system of equations?

Maru [420]

3x - y = 9
4x + y = 5
---------------add
7x = 14
x = 14/7
x = 2

3x - y = 9
3(2) - y = 9
6 - y = 9
-y = 9 - 6
-y = 3
y = -3

C. (2,-3) is a solution to the system of linear equations

4 0

3 years ago

Read 2 more answers