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Licemer1 [7]
3 years ago
11

Helpndmdkdkdndnfndndkdkdmd

Mathematics
1 answer:
alekssr [168]3 years ago
6 0

I read this story last year for my 7th grade year. The correct answer is A!

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Please help me on the question on the bottom
ratelena [41]
Answer: 4

Step -by-step :
7+7+7=21
5+12+a=
17+a
21-17=4
4 0
3 years ago
Read 2 more answers
Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15
Digiron [165]

\huge \boxed{\mathfrak{Question} \downarrow}

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{array}  { l  }  { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.

  • To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

2x-5y=-13, \: 3x+4y=15

  • Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

2x-5y=-13

  • Add 5y to both sides of the equation.

2x=5y-13

  • Divide both sides by 2.

x=\frac{1}{2}\left(5y-13\right)  \\

  • Multiply \frac{1}{2}\\ times 5y - 13.

x=\frac{5}{2}y-\frac{13}{2}  \\

  • Substitute \frac{5y-13}{2}\\ for x in the other equation, 3x + 4y = 15.

3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15  \\

  • Multiply 3 times \frac{5y-13}{2}\\.

\frac{15}{2}y-\frac{39}{2}+4y=15  \\

  • Add \frac{15y}{2} \\ to 4y.

\frac{23}{2}y-\frac{39}{2}=15  \\

  • Add \frac{39}{2}\\ to both sides of the equation.

\frac{23}{2}y=\frac{69}{2}  \\

  • Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

\large \underline{ \underline{ \sf \: y=3 }}

  • Substitute 3 for y in x=\frac{5}{2}y-\frac{13}{2}\\. Because the resulting equation contains only one variable, you can solve for x directly.

x=\frac{5}{2}\times 3-\frac{13}{2}  \\

  • Multiply 5/2 times 3.

x=\frac{15-13}{2}  \\

  • Add -\frac{13}{2}\\ to \frac{15}{2}\\ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

\large\underline{ \underline{ \sf \: x=1 }}

  • The system is now solved. The value of x & y will be 1 & 3 respectively.

\huge\boxed{  \boxed{\bf \: x=1, \: y=3 }}

8 0
2 years ago
PLZ dont ignore me! I need help on this :(
uysha [10]
Answer: It’s the second one!

Explanation: The dot is open and pointing to the side of the numbers greater than -3. It has to be an open dot because it is “greater than -3” not “greater than or equal to-3”! :)
4 0
3 years ago
b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.
Ainat [17]

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

<u>Step-by-step explanation:</u>

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ 2x

The third box contained two more pounds than the first box i.e.

⇒ x+2

The last box contained one-fourth the amount in the second box i.e.

⇒ (\frac{1}{4})2x = \frac{x}{4}

Therefore , Sum of pounds of candy are :

⇒ \frac{x}{2} +x+2+2x+x=65

⇒ \frac{x}{2} +4x=63

⇒ \frac{9x}{2}=63

⇒ x=63(\frac{2}{9} )

⇒ x=14

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

8 0
3 years ago
What is the volume of the prism below? 12 in 16 in 9 in
Inga [223]

Answer:

1728 units cubed

Step-by-step explanation:

V = LWH
V = 12x16x9
V = 1728

5 0
2 years ago
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