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3 years ago

a truck weighs 5,400 pounds. An open-wheel race car weighs 1/4 as much. how much does the race car weigh

Mathematics

2 answers:

Natasha2012 [34]3 years ago

6 0

The race car weighs 1350 pounds because 5400*one fourth is 1350

gtnhenbr [62]3 years ago

3 0

Since 1/4 is equal to 0.25 than you would do 5,400 times 0.25 and get the answer of 13,500.0. So the race car weighs 13,500 pounds.

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Nataly [62]

Answer:

y = 4x + 14

Step-by-step explanation:

slope-intercept form: y = mx + b

Slope formula: $\frac{y2-y1}{x2-x1}$

To write the equation in y = mx + b form, we need to find the slope(m) and the y-intercept(b) of the equation.

To find the slope, take two points from the table(in this example I'll use points (0, 14) and (1, 18)) and input them into the slope formula:

$\frac{18-14}{1-0}$

Simplify:

18 - 14 = 4

1 - 0 = 1

$\frac{4}{1}=4$

The slope is 4.

To find the y-intercept, input the values of the slope and one point(in this example I'll use point (1, 18)) into the equation format and solve for b:

y = mx + b

18 = 4(1) + b

18 = 4 + b

14 = b

The y-intercept is 14.

Now that we know the slope and the y-intercept, we can write the equation:

y = 4x + 14

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3 years ago

Find the missing side lengths. Answers are in simplest radical form with the denominator rationalized

Roman55 [17]

Answer:

Option B.

Step-by-step explanation:

The given triangle is a right angle triangle.

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

In the given triangle,

$\tan (45^{\circ})=\dfrac{v}{7}$

$1=\dfrac{v}{7}$

$7=v$

Using Pythagoras theorem, we get

$hypotenuse^2=Perpendicular^2+Base^2$

$u^2=v^2+7^2$

$u^2=7^2+7^2$

$u^2=2(7^2)$

Taking square root on both sides, we get

$u=\sqrt{2(7^2)}$

$u=7\sqrt{2}$

Therefore, the correct option is B.

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3 years ago

Use the diagram on the right. How many planes contain both the line and the point

Aloiza [94]

Part (a)

<h3>Answer: 0</h3>

-------------------

Explanation:

Point P is part of 3 planes or faces of this triangular prism:

plane PEF (the front slanted plane)
plane PEH (the left triangular face)
plane PHG (the back rectangular wall)

Notice how each three letter sequence involves "P", though this isn't technically always necessary. I did so to emphasize how point P is involved with these planes.

Each of the three planes mentioned do not involve line FG

Plane PEF only deals with point F
Plane PEH doesn't have any of F or G involved
plane PHG only involves G

So there are no planes that contain line FG and point P.

==================================================

Part (b)

<h3>Answer: 0</h3>

-------------------

Explanation:

It's the same idea as part (a) earlier. The planes involving point G are

plane GQF (triangular face on the right)
plane GFE (bottom rectangular floor)
plane GHP (back rectangular wall)

None of these planes have line EP going through them.

As an alternative, we could reverse things and focus on all of the planes connected to line EP. Those 2 planes are

plane PEH (triangular face on the left)
plane PEF (front slanted rectangular face)

None of these planes have point G located in them.

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Y_Kistochka [10]

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Phoenix [80]

Answer:

180 - 48 = 132

132 ÷ 2= 66

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