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Paha777 [63]
3 years ago
13

‼️help please.‼️What is the slope-intercept equation of the line with a slope

Mathematics
1 answer:
Luden [163]3 years ago
7 0

Answer:

C

Step-by-step explanation:

according to the standard equation;

Y=mX+C

or, Y=4/5X +3

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Sam rides his bike at a rate of 15 miles per hour. How many feet per minute is this?
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Si tu prend un chien de 12k je crois que ta réponse va donner 15000
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3 years ago
if rectangle ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180 degrees, where would point A lie. ​
cupoosta [38]

Answer: Point A would be at (1,-1)

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3 years ago
Where to put the ( )’s 5+4+3 2=29
Salsk061 [2.6K]

Answer:

It all depends.

Step-by-step explanation:

It depends on what you are trying to do, and what you want to solve for. If you are trying to distribute the 3 into 2, you would place the parentheses around the 2. I hope I was able to help, but there is not much that I can do without more clear instructions!

8 0
3 years ago
Simplify the expression <br> 7y+3x+3-2y+6
pishuonlain [190]

Answer:

5y + 3x + 9

Step-by-step explanation:

Whenever something is multiplied by the same variable, you can combine the coefficients. In this case, first you'd move the values with the same variable next to each other, for simplicity:

7y-2y+3x+3+6

Then, combine like terms.

5y+3x+3+6

5y+3x+9

8 0
4 years ago
A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the
kramer

Answer:

18.73ft^3

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top=(side)^2=x^2

Area of one side face=l\times b=xy

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box=V=lbh=x^2y

Total cost=9(x^2)+5x^2+4(4xy)=14x^2+16xy

204=14x^2+16xy

204-14x^2=16xy

y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}

Substitute the values of y

V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)

Differentiate w.r.t x

V'(x)=\frac{1}{8}(102-21x^2)=0

V'(x)=0

\frac{1}{8}(102-21x^2)=0

102-21x^2=0

102=21x^2

x^2=\frac{102}{21}=4.85

x=\sqrt{4.85}=2.2

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

V''(x)=\frac{1}{8}(-42x)

Substitute the value

V''(2.2)=-\frac{42}{8}(2.2)=-11.55

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value  of x

y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft

Greatest volume of box=x^2y=(2.2)^2\times 3.87=18.73 ft^3

5 0
4 years ago
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