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3 years ago

—7times(—8) help plz

Mathematics

2 answers:

BabaBlast [244]3 years ago

8 0

Answer:

Step-by-step explanation:

A negative times a negative makes a positive, so:

$-7*(-8)=56$

PilotLPTM [1.2K]3 years ago

8 0

Answer:

$56$

Step-by-step explanation:

$-7*(-8)$

The only one thing we have to do on this question is simplify.

$(-7)(-8)$

And this equals

$56$

Now you got your answer!

Hope this helps!

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Find the area of the following shape. You must show all work to receive credit. (10 points) Figure ABCDEF is shown.

andrey2020 [161]

Answer:

Step-by-step explanation:

1) Take the area of the rectangle from origin to Point B. We see that it is 8 * 11 = 88 units of area

2) We subtract all the triangles in the rectangle that are not covered by the shape.

Triangle FAPoint(0,-8) area = 1/2 * 2 * 5 = 5
Triangle FEPoint(0,-3) area = 1/2 * 3 * 4 = 6
Rectangle P(0,0) P(0,-3) P(6,-3) P(6,0) area = 6 * 3 = 18
Triangle DP(6,0)P(11,0) area = 1/2 * 5 * 3 = 7.5

Take the area of the big imaginary rectangle and subtract the area of all the smaller triangles and the smaller rectangle

88 - 5 - 6 - 7.5 - 18 = 51.5

Done!

5 0

3 years ago

Read 2 more answers

U = { z | z is an integer and − 1 ≤ z < 2 }

Effectus [21]

Answer:

(-1,0,1,2)

Step-by-step explanation:

in listing the values of z it will now be (z:z= -1,0,1,2)

3 0

2 years ago

I need help with these two math problems it’s homework and due and there provided in the picture above that being 2 and 4

Vesnalui [34]

Answer: 268.8

Step-by-step explanation: i know 2 but not 4

proportion: 5= 112

12= x

x= 268.8

4 0

3 years ago

If a < 0 then the graph opens_____

telo118 [61]

Answer:

down

Step-by-step explanation:

Assuming this is for a parabola

a>0 the graph opens up

a < 0 the graph opens down

( negative is a frown)

4 0

3 years ago

Read 2 more answers

Number 1d please help me analytical geometry

lesantik [10]

For a) is just the distance formula

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
% (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}$
-----------------------------------------------------------------------------------------
for b) is also the distance formula, just different coordinates and distance

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
% (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}$
--------------------------------------------------------------------------
for c) well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}$

now.. whatever that is, is = BC, so the distance for BC is

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
% (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
% distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}$

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d) we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
% (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\
$

$\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}$

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"

7 0

3 years ago