Question:

Solve -5(3n + 4) = 40.

Sveta_85 [38]

N=-4
Distribute the -5
-15n-20=40
-15n=60
n=-4

3 0

2 years ago

Help! Ill mark you as brain! 15 points!

Alina [70]

Answer:

35 in²

Step-by-step explanation:

The irregular shaped can be divided into two squares and one rectangle, so the area will be the additions of the area of the squares and the rectangle

area of square A = L *B = 3 *3 = 9in²

area of square B = L * B = 4*4 = 16in²

area of rectangle C = L * B = 5 *2 = 10in²

th area of the irregular shape = 9 in² + 16 in² + 10 in² = 35 in²

4 0

3 years ago

First the first 4 term for 4n-5

likoan [24]

Answer:

-1, 3, 7, 11

Step-by-step explanation:

First, let's find term 1. To find this term, we need to replace n with 1 - what is 4(1) - 5? Well, that's just 4 - 5, or -1. So, our first term is -1. Next, we'll find our second term. To do this, we'll replace n with 2. 4(2) = 8, and then 8 - 5 is equal to 3, so our second term is 3. Let's now find the third term by replacing n with the number 3. 4(3) - 5 is equal to 12 - 5, or 7. So, 7 is our third term. Finally, we'll find the fourth term. 4(4) is equal to 16, and 16 - 5 is equal to 11. So, the fourth and final term is equal to 11.

Hopefully that's helpful! :)

5 0

2 years ago

Read 2 more answers

Help me with question a please ! With full workings !

frosja888 [35]

A)

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
P&({{ 0.5}}\quad ,&{{ 0}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}$

$\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}
\\\\\\
QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}
\\\\\\
QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}$

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)

c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
% (c,d)
R&({{ -0.5}}\quad ,&{{ 0}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}$

$\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

7 0

3 years ago

If x > 0 and y > 0, where is the point (x, y) located?

Scrat [10]

Given, x > 0, y > 0.
So, the values of x and y are less than 0, that means, they are negative integers.
If we divide a Cartesian plane with the x-axis and y-axis, we get four quadrants.
1st Quadrant : (+,+)
2nd Quadrant : (-,+)
3rd Quadrant : (-,-)
4th Quadrant : (+,-)
Since the values of x and y are both negative, so (x, y) is located in the 3rd Quadrant.

Hope you could get an idea from here.

Doubt clarification - use comment section

8 0

2 years ago

￼ Question:

Question: