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3 years ago

Which expressions are equivalent to 10a - 25 + 5b

Mathematics

2 answers:

galben [10]3 years ago

5 0

Answer:

(-2a +5 - b)(-5)

Step-by-step explanation:

Liono4ka [1.6K]3 years ago

4 0

Answer:Move −25-25.

10a+5b−25

Step-by-step explanation:

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Solve the following equation for a. be sure to take into account whether a letter is capitalized or not g = Ha + na

dimulka [17.4K]

Answer:

$a = \frac{g}{H+n}$

Step-by-step explanation:

Step 1: Factor

g = a(H + n)

Step 2: Divide both sides by expression in parenthesis

g/(H + n) = a

7 0

4 years ago

Can anyone help? Find the measure of L

Sauron [17]

It’s 90 degrees hope this helps

7 0

2 years ago

Need brief explanation about why false is correct

anyanavicka [17]

We are given the equation:

(a + b)! = a! + b!

Testing the given equation

In order to test it, we will let: a = 2 and b = 3

So, we can rewrite the equation as:

(2+3)! = 2! + 3!

5! = 2! + 3!

We know that (5! = 120) , (2! = 2) and (3! = 6):

120 = 2 + 6

We can see that LHS ≠ RHS,

So, we can say that the given equation is incorrect

6 0

3 years ago

Can someone solve this question to me.i will mark brainliest. :)

Alex787 [66]

Answer:

Parent Function: $y=\sqrt{x}$
Horizontal shift: right 3 units
Vertical shift: up 3 units
Reflection about the x-axis: none
Vertical strech: streched

Step-by-step explanation:

assume that $y=\sqrt{x}$ is $f(x)=\sqrt{x}$ and $y=\sqrt{-2x+6}+3$ is

$g(x)=\sqrt{-2x+6}+3\\ f(x)=\sqrt{x} \\g(x)=\sqrt{-2x+6}+3$

The transformation from the first equation to the second equation can be found by finding a,h and k for each equation.

$y=a\sqrt{x-h}+k$

factor a 1 out of the absolute value to make the coefficient of x equal to 1

$y=\sqrt{x}$

factor a 2 out of the absolute value to make the coefficient of x equal to 1

$y=\sqrt{2}\sqrt{x-3}+3$

find a, h and k for $y=\sqrt{2}\sqrt{x-3}+3$

$a=1.41421356\\ h=3\\k=3$

the horizontal shift depends on the value of h when $h > 0$ , the horizontal shift is described as:

$g(x)=f(x+h)$ - the graph is shifted to the left h units

$g(x)=f(x-h)\\$ - the graph is shifted to the right h units

the vertical shift depends on the value of k

5 0

2 years ago

100 POINTS TO RIGHT AWNSER!!!!

makvit [3.9K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586

——————————

The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D) r < 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

Now all real numbers are included in the union. So this is the right choice.

Answer: option D) r < 5 or r > – 1.

I hope this helps. =)

6 0

3 years ago