MATHEMATICS MIDDLE SCHOOL

Bryan played two baseball games last weekend. He got 7 hits in all.He had 3 hits in the first game. How many hits did he get in the
second game?

Answers

Answer 1
Answer:

Answer: It's probably 4.

Step-by-step explanation:

7-3=4
Answer 2
Answer:

Answer:

4 hits in the second game

Step-by-step explanation:

If he hit 3 in the first game and 7 total, then the remainder (7-3) would be 4 for the second game,

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12x−5y=−20
y=x+4
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Answers

Answer:

12x-5(x+4)=-20

7x=0

x=0

so,y=4
ANSWER:

x=0

STEP BY STEP EXPLANATION:

1.) 12x-5y=-20
y=x+4

2.) 12x-5(x+4)=-20
*because you are taking the y= equation and putting that into 5y replacing the y.

3.) 12x-5x-20=-20
*because you multiply -5 by the x and the 4 and get -5x-20.

4.) 7x-20=-20
*because you subtract 5x from 12x because you are using like terms to make the equation smaller.

5.) 7x=0
*because you add the 20 on both sides to cancel out the 20 to get x closer to being by itself.

6.) x=0
*because you divide 7 on both sides to get x completely by itself.

Find the slope of the line described by 3x + 2y + 1 = 0. 3/2
−2/3
−3/2
−1/2

Answers

3x + 2y + 1 = 0

subtract 3x+1 from each side

3x + 2y + 1  - 3x -1=  -3x -1

2y = -3x-1

divide by 2 on each side

2y/2 = -3x/2 -1/2

y = -3/2 x -1/2

this is in the form y = mx +b where m is the slope

m = -3/2

the slope is -3/2

solve system by substitution

y=3x-20

y=-x^(2)+34

Answers

\begin{cases} y=3x-20 \n y=-x^2+34 \end{cases}\n \n \begin{cases} y=3x-20 \n3x-20=-x^2+34 \end{cases}\n \n\begin{cases} y=3x-20 \nx^2+3x-20-34=0 \end{cases}\n \n\begin{cases} y=3x-20 \nx^2+3x-54=0 \end{cases}

\begin{cases} y=3x-20 \nx^2+3x-54=0 \end{cases} \n \nx^2+3x-54=0 \n \na=1, \ \ b=3, \ \ c= -54 \n \n \Delta =b^2 -4ac =3^2-4*1*(-54)=9+216=225 \n \nx_(1)=(-b-√(\Delta ))/(2a) =(-3-√(225))/(2)=(-3-15)/(2)= (-18)/(2)=-9

x_(2)=(-b+√(\Delta ))/(2a) =(-3+√(225))/(2)=(-3+15)/(2)= (12)/(2)=6

\begin{cases} y=3x-20 \n x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=3x-20 \nx=6 \end{cases}\n \n \begin{cases} y=3*(-9)-20 \n x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=3*6-20 \nx=6 \end{cases}\n \n \begin{cases} y=-27-20 \n x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=18-20 \nx=6 \end{cases}\n \n \begin{cases} y=-47 \n x=-9 \end{cases}\ \ \vee \ \ \begin{cases} y=-2\nx=6 \end{cases}

15) 3x2 y4 z − 6xy − 3z(5x2) How many terms are in the expression?a. 1
b. 2
c. 3
d. 4

Answers

Ok name the expressions we both know 3z is one 5x is one 6xy is one do you think 3x^2y^4z is an expression i would give you the answer but i also want you to understand the problem and what your looking for to in the equation

How do I find the missing measurements?

Answers

Answer:

CD = 10

AD = 22

EC = 13

AC = 26

∠CAD = 23°

∠BAC = 47°

∠ADC  =110°

Step-by-step explanation:

A parallelogram is a quadrilateral with two sets of opposite parallel lines.

Which mean in ABCD parallelogram given AB is parallel to DC and BC is parallel to AD.

Now lets find the values required,

CD = AB (Opposite sides are equal in length)

But, AB= 10 (given) therefore CD = 10.

AD = BC (Opposite sides are equal in length)

But, BC= 22 (given) therefore AD = 22.

EC = AE (Diagonals bisect each other)

But, AE= 13 (given) therefore EC = 13

AC = AE + EC

= 13+13

AC =26

∠CAD = ∠BCA (Alternate interior angles are equal)

But, ∠BCA = 23° therefore, ∠CAD = 23°

∠BAC = ∠ACD (Alternate interior angles are equal)

But, ∠ACD = 47° therefore, ∠BAC = 47°

∠ADC + ∠BCD = 180° (Consecutive interior angles)

But,  ∠BCD = 23°+47°=70°

∠ADC = 180°- ∠BCD

           =180°-70°

∠ADC  =110°

please please please please help what is the area of trapezoid ABCD enter your answer as a decimal or whole number do not round at any steps help

Answers

Answer:

A = 62.5

Step-by-step explanation:

I would break this up into two easier cases. the square at the top where three vertices are A,B, and C. and then the triangle for the rest of the area. we can use the distance formula to find the side length of the square.

sqrt((y2-y1)^2 + (x2-x1)^2)

sqrt((5-2)^2 + (-1-3)^2)

sqrt(3^2 + (-4)^2)

sqrt(9 + 16)

sqrt ( 25)

5

so the area of the square is 5^2 =25

now for the triangle. the formula for the area of a triangle is (1/2)bh.

so we can plug in our b=5, but we need to find h.

we use the distance formula between point A and D to find the distance to be 20, but we have to subtract the 5 we already counted in the area of the square. so the height is 15.

so The triangle has area (1/2)×5×15

37.5.

add the two areas together and we get our answer

25+37.5 = 62.5
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