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3 years ago

Match the equation with its corresponding solution for x. 4x + 7 = 35

1 answer:

8 0

Are you wanting to solve for x? If so:
4x + 7 = 35
We want to isolate the variable, so subtract 7 from both sides. You will get:
4x = 28
Now, divide each side by 4 to get X by alone. This will give you:
x = 7
I hope this helped.

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Find the following when a = -1, b = 2, c = -1/4<br> (4b - 3)/(2a) -1

Katena32 [7]

We substitute the given into the equation
given: a=-1, b=2, c= $\frac{-1}{4}$
=( $\frac{4b-3}{2a}$ -1
= $\frac{4(2)-3}{2(-1)}$ -1
= $\frac{8-3}{-2}$ -1
= $\frac{5}{-2}$ -1
to continue, we can either find a common denominator or change the fraction to decimal.
Fraction to Decimal:
$\frac{5}{-2}$ =-2.5
Equation=-2.5-1=-3.5
OR Common Denominator:
= $\frac{5}{-2}$ $\frac{-1}{1}$ × $\frac{-2}{-2}$ (denominator should be -2=common)
= $\frac{5}{-2}$ + $\frac{2}{-2}$
= $\frac{7}{-2}$
which also = to -3.5

3 0

3 years ago

Evaluate the expression 3 ⋅ 2^t , when t=1. *

lisov135 [29]

Answer:

Step-by-step explanation:

3 ⋅ 2^t

Let t=1

3 ⋅ 2^1

Exponents first

3 * 2

8 0

3 years ago

Help pls shejsbsjsjshsjsvshzhabjs

alex41 [277]

I think it’s 1. If wrong sorry

8 0

2 years ago

Read 2 more answers

If 2^n + 1 is an odd prime for some integer n, prove that n is a power of 2. (H

vovikov84 [41]

Step-by-step explanation:

We will prove by contradiction. Assume that $2^n + 1$ is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that $p\mid n$ . Then, for some integer $k\geq 1$ ,

$n=p\times k.$

Therefore

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.$

Here we will use the formula for the sum of odd powers, which states that, for $a,b\in \mathbb{R}$ and an odd positive number $n$ ,

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})$

Applying this formula in 1) we obtain that

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p=(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^{k}+1)$ .

Then, as $2^k+1>1$ we have that $2^n+1$ is not a prime number, which is a contradiction.

In conclusion, if $2^n+1$ is an odd prime, then n must be a power of 2.

4 0

3 years ago

How do I solve this?

kramer

Https://www.symbolab.com/solver/step-by-step/%5Ccot%5Cfrac%7B-5%5Cpi%7D%7B6%7D

I'm unsure how to properly express how to do this to you but here is a list of steps on how to solve it. Good luck.

7 0

3 years ago