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Aliun [14]
3 years ago
10

There was 2/3 of a fruitcake left over. Ella ate 1/5 of it. How much of the whole fruitcake did she eat?

Mathematics
1 answer:
liq [111]3 years ago
7 0
2/3 * 1/5 = 2/15
A) 2/15
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HELP I NEED SOMBODEY HELP NOT JUST ANYONE JK HELP ME PLSSSSS
tatuchka [14]
You cross multiply/butterfly method so 5•6= 30 and 5•8=40 so it would be 30/40 OR 3/4 simplified
8 0
3 years ago
A particle moves with velocity function v(t) = 2t^2 - 3t - 3, with v measured in feet per second and t measured in seconds. Find
SashulF [63]
V(t) = 2t^2 - 3t - 3

for an acceleration a(t),
a(t) = d (v(t)) / d t
a(t) = 4t - 3

at t=2
a(2) = 8 - 3
a(2) = 5 feet/s^2

3 0
3 years ago
Which is the answer and how did you solve it?
motikmotik
Its pretty simple

9x-7y = 22
x + 3y = -24

(9x-7y = 22)
-9(x + 3y = -24)
= -9x -27y = 216
+ 9x-7y = 22
= -34y = 238
= / -34 = 238/-34
y = -7

now plug -7 in one of the equation
x + 3(-7) = -24
x -21 = -24
+21. +21
x = -3

so answer is x = -3 and y = -7

3 0
3 years ago
When 3a + 7b > 2b -8b is solve for a the result is
Travka [436]

Answer:

3a+7b>2b-8b

3a>2b-8b-7b

3a>2b-15b

3a>-13b

a>-13b/3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
3. Try It #3 Write the point-slope form of an equation of a line with a slope of -2 that passes through the point (-2,2). Then r
vova2212 [387]

Answer:

Point-slope form of equation given as $y-2=-2(x+2)$.

Slope-intercept form of equation is given as $y=-2 x-2$.

Step-by-step explanation:

In the question, it is given that the slope of a line is -2 and it passes from (-2,2).

It is asked to write the point-slope form of the equation and rewrite it as slope-intercept form.

To do so, first find the values which are given in the question and put it in the formula of point-slope form. Simplify the equation to rewrite as slope-intercept form.

Step 1 of 2

Passing point of the line is (-2,2).

Hence, $x_{1}=-2$ and

$$y_{1}=2 \text {. }$$

Also, the slope of the line is -2.

Hence, m=-2

Substitute the above values in point-slope form of equation given by $y-y_{1}=m\left(x-x_{1}\right)$

$$\begin{aligned}&y-y_{1}=m\left(x-x_{1}\right) \\&y-2=-2(x-(-2) \\&y-2=-2(x+2)\end{aligned}$$

Hence, point-slope form of equation given as y-2=-2(x+2).

Step 2 of 2

Solve y-2=-2(x+2) to write it as slope-intercept form given by y=mx+c.

$$\begin{aligned}&y-2=-2(x+2) \\&y-2=-2 x-4 \\&y=-2 x-4+2 \\&y=-2 x-2\end{aligned}$$

Hence, slope-intercept form of equation is given as y=-2x-2.

7 0
2 years ago
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