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3 years ago

Suppose that 16 inches of wire costs 48 cents. At the same rate, how many inches of wire can be bought for 33 cents?

2 answers:

7 0

The answer is 11.
if 16 inches of wire cost 48 cents your goal i to figure out how many inches of wire cots 33 cents to do so you have to find out how many cents each inch is. to do that you have to divide 48 by 16 and you get 3 so that mean that 1 inch costs 3 cents. after you figure that out you do 33 divided by 3 and you get 11. the answer is 11 inches costs 33 cents.

LenaWriter [7]3 years ago

3 0

11

16/48=0.33333333333
0.33333333333*33=11

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Y=mx+b is used for what math equation

soldier1979 [14.2K]

y = m x + b is a linear equation.
This equation is in the slope - intercept form, where m is the slope and b is y - intercept.
For example: y = 2 x + 5
In this equation: m = 2 ( the slope ) and b = 5 ( the line intercepts y-axis at the point ( 0, 5 ) ).

3 0

3 years ago

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3 is less than or equal to 7+g

Fantom [35]

2≤4 so you have to do that by seeing that 3 iss less that 7

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4 years ago

In Exercise 4, find the surface area of the solid<br> formed by the net.

Fittoniya [83]

Answer:

3. 150.72 in²

4. 535.2cm²

Step-by-step Explanation:

3. The solid formed by the net given in problem 3 is the net of a cylinder.

The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.

The surface area = Area of the 2 circles + area of the rectangle

Take π as 3.14

radius of circle = ½ of 4 = 2 in

Area of the 2 circles = 2(πr²) = 2*3.14*2²

Area of the 2 circles = 25.12 in²

Area of the rectangle = L*W

width is given as 10 in.

Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in

Area of rectangle = L*W = 12.56*10 = 125.6 in²

Surface area of net = Area of the 2 circles + area of the rectangle

= 25.12 + 125.6 = 150.72 in²

4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)

= $2(0.5*b*h) + 3(l*w)$

Where,

b = 8 cm

h = $\sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A = 2(0.5*8*6.9) + 3(20*8)$

$S.A = 2(27.6) + 3(160)$

$S.A = 55.2 + 480$

$S.A = 535.2 cm^2$

6 0

4 years ago

Can someone tell me if im right? PLEASE DONT JUST SAY YES if you don’t know

tia_tia [17]

Answer: You are incorrect, the slope is correct, but the actual y-intercept is 205 ft.

Then the equation is:

y = (-15 ft/min)*x + 205 ft

Step-by-step explanation:

Ok, let's solve this.

We know that water is drained from a reservoir, let's assume that we can model this situation with a linear relation:

y = a*x + b

Where x is time, y is the height of the water in the reservoir, a is the slope (in this case represents how much changes the height of the water in the reservoir in one unit of time) and b is the initial height of the water in the reservoir.

We know that for a line that passes through the points (x₁, y₁) and (x₂, y₂) the slope is:

a = (y₂ - y₁)/(x₂ - x₁)

For this particular case we know that after 2 minutes the height of the water is 175 ft, then we have the point (2 min, 175 ft)

and after 5 minutes (so 7 minutes in total), the height of the water is 100ft, then: (7 ft, 100ft)

Then the slope of this:

a = (100 ft - 175 ft)/(7 min - 2 min) = (-75ft/5min) = - 15 ft/min

Then our line is something like:

y = (-15ft/min)*x + b

To find the value of b, we can use the fact that when x = 2 min, y = 175 ft

So if we replace these two values in the equation we get:

175ft = (-15 ft/min)*2 min + b

175 ft = -30 ft + b

175 ft + 30 ft = b

(here is your problem, it seems like you subtracted instead of adding in this part)

205 ft = b

Then the equation is:

y = (-15 ft/min)*x + 205 ft

So you are incorrect (but only for a little bit), you computed wrong the y-intercept.

3 0

3 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, –5). Describe the ste

olga_2 [115]

1. "the graph has the same zeros" : so let a be the "triple" root of the cubic polynomial function.

2. So f(x)= $(x-a)^{3}$

3. Don't forget that the expression might have a coefficient b as well, and still maintain the conditions:

$f(x)=b(x-a)^{3}$

4. Now, f(0)=-5 so $-5=f(0)=b(0-a)^{3}=b(-a) ^{3}=-ba ^{3}$

$-5=-ba ^{3}$

$5=ba ^{3}$

$b= \frac{5}{ a^{3} }$

5. the function is $f(x)=\frac{5}{ a^{3} }(x-a)^{3}$ where a can be any real number except 0

7 0

4 years ago

Read 2 more answers