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Vikentia [17]
3 years ago
14

—7times(—8) help plz

Mathematics
2 answers:
BabaBlast [244]3 years ago
8 0

Answer:

56

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:

53

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.

So

  • 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 \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ x}}\quad ,&{{ 1}})\quad &#10;%  (c,d)&#10;B&({{ -4}}\quad ,&{{ 1}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}&#10;\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -7}}\quad ,&{{ y}})\quad &#10;%  (c,d)&#10;B&({{ -3}}\quad ,&{{ 4}})&#10;\end{array}\ \ &#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}&#10;\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 \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -3}}\quad ,&{{ 0}})\quad &#10;%  (c,d)&#10;B&({{ 5}}\quad ,&{{ -2}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\&#10;d=\boxed{?}&#10;&#10;\end{array}

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

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;B&({{ 5}}\quad ,&{{ -2}})\quad &#10;%  (c,d)&#10;C&({{ -13}}\quad ,&{{ y}})&#10;\end{array}\qquad &#10;%  distance value&#10;\begin{array}{llll}&#10;&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\&#10;d=BC\\\\&#10;BC=\boxed{?}&#10;&#10;\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 \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;M&({{-2}}\quad ,&{{ 1}})\quad &#10;%  (c,d)&#10;N&({{ x}}\quad ,&{{ 1}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P&#10;\\\\\\&#10;

\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies &#10;\begin{cases}&#10;\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\&#10;\cfrac{{{ y_2}} + {{ y_1}}}{2}=4&#10;\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
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