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3 years ago

If you help I got more questions for you and thank you answering help needed now

Mathematics

1 answer:

Grace [21]3 years ago

4 0

Answer:

13,9,5

Step-by-step explanation:

The problem keeps taking away 4 from the previous number.

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I forgot how to convert the inequalities so if someone could please help me that would be great!

jeyben [28]

WHEN GRAPHING

Step 1. Convert to y = mx + b

Equation 1: y = 3x/2 - 16/2
Equation 2: y = -5 - 2x

Step 2. Graph like normal

Find zeros & plot
Find y-intercept & plot

Step 3. Shade as indicated by the inequality symbol

> or ≥ = above line
< or ≤ = below line

Step 4. If ≤ or ≥ ONLY, then also shade the line

FOR THIS PROBLEM

1. Graph each equation

2. Shade ABOVE line for each

3. Shade line first equation as well

Hope this helps and God bless!

4 0

2 years ago

What is the true solution to the equation below?

Marina CMI [18]

Answer:

The solution is:

$x=4$

Step-by-step explanation:

Considering the expression

$lne^{lnx}+lne^{lnx}^{^2}=2ln8$

$\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)$

$\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b$

$\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2$

$\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)$

$\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)$

$3\ln \left(x\right)=2\ln \left(8\right)$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}$

$\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A$

Solving the right side of the equation A.

$\frac{2\ln \left(8\right)}{3}$

$\ln \left(8\right):\quad 3\ln \left(2\right)$

Because

$\ln \left(8\right)$

$\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3$

⇒ $\ln \left(2^3\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^3\right)=3\ln \left(2\right)$

$\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6$

$=\frac{6\ln \left(2\right)}{3}$

$\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2$

$=2\ln \left(2\right)$

So, equation A becomes

$\ln \left(x\right)=2\ln \left(2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$=\ln \left(2^2\right)$

$=\ln \left(4\right)$

$\ln \left(x\right)=\ln \left(4\right)$

$\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)$

$x=4$

Therefore, the solution is

$x=4$

6 0

3 years ago

Read 2 more answers

Which is the equation of a line that passes through (1,8) and (6,4)

Sati [7]

Answer:

y = - $\frac{4}{5}$ x + $\frac{44}{5}$

Step-by-step explanation:

(6, 4)

(1, 8)

m = $\frac{8-4}{1-6}$ = - $\frac{4}{5}$

y - 4 = - $\frac{4}{5}$ ( x - 6 )

y = -  $\frac{4}{5}$  x +  $\frac{44}{5}$ ⇔ y = - 0.8x + 8.8

8 0

3 years ago

For all values of x. which expression is equivalent to (3x + 2) + 2(3x +2)

iris [78.8K]

Answer:

9x+6

Step-by-step explanation:

(3x+2)+2(3x+2)

3x+2+6x+4

9x+6

3 0

3 years ago

23.

Crazy boy [7]

Answer:

The number that cannot be the largest possible 6-digit number is;

(D) AAABCB

Step-by-step explanation:

From the question, we have;

A, B, and C = Distinct digits, therefore, A ≠ B ≠ C

The number of digits in the number to be formed = 6 digits

The number of 'A' in the number to be formed = 3

The number of 'B' in the number to be formed = 2

The number of 'C' in the number to be formed = 1

We have;

When A > B > C

The largest possible number = AAABBC

When C > A > B

The largest possible number = CAAABB

When B > A > C

The largest possible number = BBAAAC

When A > C > B

The largest possible number = AAACBB

Therefore, given that when A > B > C, the largest possible number = AAABBC, we have;

AAABBC > AAABCB, because B > C, therefore, within the tens and unit of the two 6 digit numbers, we have, BC > CB

∴ AAABBC > AAABCB and AAABCB, cannot be the largest possible 6-digit number

3 0

3 years ago