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3 years ago

Solve. 5 = –(–z + 3) a. –8 b. 8 c. –2 d. 2

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

6 0

-(-z + 3) equals z - 3

5 = z - 3

z = 8

gulaghasi [49]3 years ago

5 0

-(-z + 3) equals z - 3
5 = z - 3
z = 8

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HELP!! find the value of x please

nevsk [136]

I am orearte sure the correct answer is 280 please give me brainliest if u can

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3 years ago

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

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3 years ago

HELPPPP- you just got your driver's license. Your parents buy you a car and agree to let you pay them back over a period of 70 m

Nitella [24]

Using the monthly payments formula, it is found that a car with a value of at most $25,293.

<h3>What is the monthly payment formula?</h3>

It is given by:

$A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}$

In which:

P is the initial amount.

r is the interest rate.

n is the number of payments.

In this problem, we have that the parameters are given as follows:

A = 400, n = 70, r = 0.035.

Hence:

r/12 = 0.035/12 = 0.002917.

Then we have to solve for P to find the maximum value of the car.

$A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}$

$400 = P\frac{0.002917(1.002917)^{70}}{(1.002917)^{70}-1}$

$P = \frac{400[(1.002917)^{70}-1]}{0.002917(1.002917)^{70}}$

P = $25,293.

More can be learned about the monthly payments formula at brainly.com/question/26267630

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2 years ago

The nutritional chart on the side of a box of a cereal states that there are 87 calories in a 3/4 cup serving. How many calories

alexira [117]

Answer:

Total calories in 8 cup = 928 calories

Step-by-step explanation:

Given:

Calories in 3/4 cup of cereal = 87 calories

Find:

Total calories in 8 cup

Computation:

Total calories in 8 cup = 8 x 87 x [4/3]

Total calories in 8 cup = 928 calories

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3 years ago

The distance d that a certain particle moves may be calculated from the expression d = at + bt2 where a and b are constants; and

pychu [463]

Guven that the <span>distance d that a certain particle moves may be calculated from the expression $d = at + bt^2$ where a and b are constants; and t is the elapsed time.

Distance is a length and hence the dimension of distance is L.

Now, $at$ and $bt^2$ also will have the dimension of L.

</span><span>Time has a dimension of T.

For $at$ , let the dimension of $a$ be $X$ , then
$XT=L \\ \\ \Rightarrow X=\frac{L}{T}$

For $bt^2$ , let the dimension of $b$ be $Y$ , then
$YT^2=L \\ \\ \Rightarrow Y= \frac{L}{T^2}$

Therefore, the dimension of $a$ is $\frac{L}{T}$ while the dimension of $b$ is $\frac{L}{T^2}$ .
</span>

4 0

3 years ago