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2 years ago

Help pls

Mathematics

1 answer:

Serga [27]2 years ago

6 0

Answer:

x=0 and y=5

Step-by-step explanation:

8x-25+2x=-25

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Find the positive base $b$ in which the equation $5_b \cdot 23_b = 151_b$ is valid.

DanielleElmas [232]

Answer:

b=7

Step-by-step explanation:

We want to determine the positive base b in which:

$5_b \cdot 23_b = 151_b$

The easiest way to approach this is to convert all the numbers to base 10.

$5_b=5Xb^0=5\\23_b =(2Xb^1)+(3Xb^0)=2b+3\\151_b=(1Xb^2)+(5Xb^1)+(1Xb^0)=b^2+5b+1\\Therefore:\\5_b \cdot 23_b = 151_b\\5(2b+3)=b^2+5b+1\\10b+15=b^2+5b+1\\b^2+5b+1-10b-15=0$

$b^2-5b-14=0$

Next, we factorize the resulting expression.

$b^2-5b-14=b^2-7b+2b-14=0\\b(b-7)+2(b-7)=0\\(b-7)(b+2)=0\\b-7=0\: or \: b+2=0\\b=7\: or \: b=-2$

The positive value of b for which the equality hold is 7.

3 0

2 years ago

What is another way to represent 100

NeX [460]

Answer:

The Roman numeral Ⅽ

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

F(x) = ax2 - 12, g(x) = 2x +3, and f(3) = 24. What<br> is the value of g(f(2)) ?

elixir [45]

<h2>Answer: g(f(2)) = 11</h2>

Step-by-step explanation:

g(f(2)) is substituting the value of f(2) for x in g(x). But we must first find f(2).

We know that f (x) = ax² - 12

Since f(3) = 24

⇒ a(3²) - 12 = 24

9 a = 36

a = 4

∴ f(2) = (4)(2²) - 12

= 4

⇒ g(f(2)) = 2(4) + 3

= 11

6 0

2 years ago

Read 2 more answers

I will mark as brainiest for correct answer! Please help me! Thank you

Dmitrij [34]

For the division part:
$\frac{5-4}{4-3 }= \frac{1}{1 }=1$

But I think your question differs than this..

Am I right?

as your options do not include 1

3 0

2 years ago

Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex

siniylev [52]

Answer:

$\large \boxed{\sf \ \ \ p=-11 \ \ \ }$

Step-by-step explanation:

Hello,

$\alpha \text{ and } \beta \text{ are the roots of the following equation}$

$2x^2+6x-7=p$

It means that

$2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\$

And we know that

$\alpha= 2\cdot \beta$

So we got two equations

$2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1$