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spayn [35]
3 years ago
10

What is the answer to 9×(6+7-3)-2​

Mathematics
2 answers:
rodikova [14]3 years ago
8 0

Answer:

88

Step-by-step explanation:

I know the answer because I used the correct method

ZanzabumX [31]3 years ago
3 0

Answer:

88

Step-by-step explanation:

PEMDAS

9*(6+7-3)-2

1) 6+7=13

13-3=10

9*10-2

2)9*10= 90

3)90-2= 88

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PLZ HELP ME WITH THIS PROBLEM I DON'T HAVE THAT MUCH TIME!!
pantera1 [17]
A. -20,-12,-8,-1,(1),5,10,x,x........ur other 2 numbers have to be greater then 1

B. x,-20,-12,-8,(-3),-1,1,5,10......ur other number has to be less then -3
8 0
3 years ago
Read 2 more answers
Help meeeeeeeeeeeeeee pleaseeeeeee
Rina8888 [55]

Answer: 2.5, 5.4

Step-by-step explanation:

-16t^2 +126t=213\\\\16t^2 -126t+213=0\\\\t =\frac{-(-126) \pm \sqrt{(-126)^2 -4(16)(213)}}{2(16)}\\\\t \approx 2.5, 5.4

8 0
1 year ago
I need help please and thank you
Anit [1.1K]

Answer:

4

Step-by-step explanation:

The slope of the line is given. It is 4. Use the equation y=mx+b to answer the other questions.

Substitute 1 in for y, 4 in for m, and 2 in for x and solve for b. The equation would look like 1=4(2)+b.

Solving that equation will give you b=-7

That means the y-intercept is -7.

The equation would then be y=4x-7

6 0
3 years ago
Nima uses 2/3 cup peanuts, 1/2 cup cashews, 3/4 cups pecans, and some raisinsin a recipe that makes 2 1/4 cups of trail mix. How
Lesechka [4]

Answer:

8/27 cups

Step-by-step explanation:

Nima uses 2/3 cups of peanuts to make a recipe of 2 1/4 = 9/4 cups of trail mix. If we want to find how many cups of peanuts are there per cup of trail mix, we can do a rule of three:

If 9/4 cups of trail mix need 2/3 cups of peanuts, 1 cup of trail mix need X cups of peanuts:

9/4 cups of trail mix -> 2/3 cups of peanuts

1 cup of trail mix -> X cups of peanuts

9/4 / 1 = 2/3 / X

9/4 = 2/(3*X)

3*X*9 = 4*2

27*X = 8

X = 8/27 cups

7 0
3 years ago
The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?
olga nikolaevna [1]
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
%  (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
%   coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2
3 0
3 years ago
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