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3 years ago

5x - 7 = -12 please help me with this ill mark brainliest

Mathematics

2 answers:

Katarina [22]3 years ago

7 0

The answer is x = -1

Leno4ka [110]3 years ago

6 0

Answer:

x=-1

Step-by-step explanation:

You might be interested in

Solve the equation for x^2= 625

docker41 [41]

Answer:

x = ±25

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Multiple Roots

Step-by-step explanation:

Step 1: Define

x² = 625

Step 2: Solve for x

Square root both sides: x = ±25

Step 3: Check

Plug in x into the original equation to verify it's a solution.

x = -25

Substitute in x: (-25)² = 625
Exponents: 625 = 625

Here we see that 625 does indeed equal 625.

∴ x = -25 is a solution to the equation.

x = 25

Substitute in x: 25² = 625
Exponents: 625 = 625

Here we see that 625 does indeed equal 625.

∴ x = 25 is a solution to the equation.

8 0

3 years ago

Suppose the base and height ( 8 and 6) are multiplied by 1/2, what effect would this have on the area

Feliz [49]

Answer:

Here is the answer hope it helps:)

6 0

2 years ago

A. The product of p and 48 How would I write this as an algebraic expression

marishachu [46]

Answer:

p * 48

Step-by-step explanation:

Lets break this down into something simpler

"The product"

answer to a multiplication problem

"of"

to multiply

"p"

an unknown value

"48"

factor

8 0

2 years ago

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

$\frac{d}{dx}(\frac{2x+3}{3x^2-4})$

$\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}$

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

6 0

3 years ago

Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given the relevant sample result

9966 [12]

Answer:

a) $\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

b) $ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

c) $(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_1$ represent the real population proportion 1

$\hat p_1=0.2$ represent the estimated proportion 1

$n_1=60$ is the sample size required 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.35$ represent the estimated proportion 2

$n_2=100$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_2)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

Part a

The best estimate is given by:

$\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

Part c

And replacing into the confidence interval formula we got:

$(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$

And the 99% confidence interval would be given (-0.319;0.0185).

4 0

3 years ago