Ask question

Ask question

All categories

3 years ago

10

Which value represents the slope of the line that has an x-intercept at (-4,0) and a y-intercept at (0, -12) in the coordinate p

lane?
1.-3
2.-1/3
3.1/3
4.3

1 answer:

Lesechka [4]3 years ago

6 0

To find the gradient you do change in y/change in x so...

-12 - 0
————- = -12/4 = -3
0 - - 4

Hope this helps! Any questions let me know :)

You might be interested in

Can someone help me please? I forgot how to do this

tensa zangetsu [6.8K]

Answer:

90°-33°=57°

57=2x+1

-1. -1

56=2x divide both sides by 2

28=x

x =28

8 0

3 years ago

Read 2 more answers

What value of c makes the equation -3c/4=24 true ?

Sonbull [250]

Creo que C es igual a -6

4 0

3 years ago

Read 2 more answers

Help with matrices please? Any wrong/not applicable answers will be reported and BLOCKED

marin [14]

m x H = $\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]$

Step-by-step explanation:

Step 1; Multiply 5 with this matrix $\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]$ and we get a matrix $\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]$

Multiply the fraction $\frac{2}{5}$ with the matrix $\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]$ and we get $\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]$

Step2; Now equate corresponding values of the matrices with each other.

-5 = $\frac{-2m}{5}$ and so on. By equating we get the value of m as $\frac{25}{2}$

Step 3; Add the matrices to get the value of matrix m.

Adding the three matrices on the RHS we get $\left[\begin{array}{ccc}2&9&-9\\\end{array}\right]$ .

Step 4; Adding the matrices on the LHS we get the resulting matrix as H +

$\left[\begin{array}{ccc}4&6&-8\\\9\end{array}\right]$ . Equating the matrices from step 3 and 4 we get the value of H as $\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]$

Step 5; Now to find the value of m x H we need to multiply the value of $\frac{25}{2}$ with the matrix $\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]$

Step 6; Multiplying we get the matrix m x H = [ -25 $\frac{75}{2}$ $\frac{-25}{2}$ ]

8 0

3 years ago

The sides of a square are 2^4/9 inches long. what is the area of the square?

bagirrra123 [75]

The Area of the square is (side)^2

side is given to be 2^(4/9)
SO,
side^2 = [2^(4/9)]^2 = 2^(4/9 * 2) = 2^(8/9)
so the area of square is 2^(8/9) sq. inches.

5 0

3 years ago

Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?.

marin [14]

To solve the problem we must know the basic exponential properties.

<h3>What are the basic exponent properties?</h3>

${a^m} \cdot {a^n} = a^{(m+n)}$

$\dfrac{a^m}{a^n} = a^{(m-n)}$

$\sqrt[m]{a^n} = a^{\frac{n}{m}}$

$(a^m)^n = a^{m\times n}$

$(m\times n)^a = m^a\times n^a$

The expression can be written as $x^9\sqrt[3]{y}$ .

Given to us

$(x^{27}y)^\frac{1}{3}$

$(x^{27}y)^\frac{1}{3}$

Using the exponential property $(m\times n)^a = m^a\times n^a$ ,

$=(x^{27}y)^\frac{1}{3}\\\\=x^{\frac{27}{3}}\times y^\frac{1}{3}\\\\=x^9\times y^\frac{1}{3}$

Using the exponential property $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ ,

$=x^9\times y^\frac{1}{3}\\\\=x^9\times \sqrt[3]{y}\\\\=x^9 \sqrt[3]{y}$

Hence, the expression can be written as $x^9\sqrt[3]{y}$ .

Learn more about Exponent properties:

brainly.com/question/1807508

5 0

2 years ago

Read 2 more answers