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2 years ago

Simplify each expression into a single number using standard algorithms. 1.-3-2

Mathematics

1 answer:

liberstina [14]2 years ago

4 0

Answer:

- 5

Step-by-step explanation:

Step 1:

- 3 - 2 Equation

Step 2:

- 3 - 2 Add

Answer:

- 5

Subtracting from a negative number is the same thing as adding. If you have a negative number and you have to subtract from it. You just add both of the negative numbers together.

Hope This Helps :)

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This morning I went to Larry's for breakfast.

Olin [163]

Answer:

9 dollars and 18 cents

Step-by-step explanation:

6 0

2 years ago

Please help! If GDEF is a square, GE=4x+1 and DF=8x-51, find the value of x.

zepelin [54]

Answer:

x = 13

Step-by-step explanation:

1. 4x + 1 = 8x - 51

2. 4x = 8x - 52

3. -4x = -52

4. x = 13

5 0

2 years ago

Read 2 more answers

Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ

const2013 [10]

Answer:

$\frac{(2a + 1)^2}{50a}$

Step-by-step explanation:

Given

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$

Required

Find the equivalent

We start by changing the / to *

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$

$\frac{2a + 1}{10a - 5} * \frac{4a^2 - 1}{10a}$

Factorize 10a - 5

$\frac{2a + 1}{5(2a - 1)} * \frac{4a^2 - 1}{10a}$

Expand 4a² - 1

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1}{10a}$

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1^2}{10a}$

Express (2a)² - 1² as a difference of two squares

Difference of two squares is such that: $a^2- b^2= (a+b)(a-b)$

The expression becomes

$\frac{2a + 1}{5(2a - 1)} * \frac{(2a - 1)(2a + 1)}{10a}$

Combine both fractions to form a single fraction

$\frac{(2a + 1)(2a - 1)(2a + 1)}{5(2a - 1)10a}$

Divide the numerator and denominator by 2a - 1

$\frac{(2a + 1)((2a + 1)}{5*10a}$

Simplify the numerator

$\frac{(2a + 1)^2}{5*10a}$

$\frac{(2a + 1)^2}{50a}$

Hence,

$\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}$ = $\frac{(2a + 1)^2}{50a}$

4 0

2 years ago

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

5 0

2 years ago

At 6 a.m. the temperature was -9oF. By 11 a.m., the temperature had increased 34oF. What was the temperature at 11 a.m.?

lesantik [10]

The resulting temperature was 25oF.

7 0

2 years ago