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Mnenie [13.5K]
3 years ago
15

HELP PLEASE!!!!!!!!!! plz

Mathematics
2 answers:
scoundrel [369]3 years ago
7 0
0       0      (0,0)
11     2     (11,2)
22     4     (22,4)
33     6     (33,6)
Anettt [7]3 years ago
7 0
If you had 0 teaspoons of powdered fertilizer, you can't make any liquid fertilizer. So the first point should be (0, 0).

Let's find out how much powdered fertilizer you need to make 1 gallon of liquid fertilizer.

11 / 2 = 5.5

So you need 5.5 teaspoons of powdered fertilizer to make 1 gallon of liquid fertilizer. This is our unit rate.

The 3rd value in the table is 22 teaspoons, it's asking how many gallons of liquid fertilizer you can make with these 22 teaspoons. We can find this out by dividing by our unit rate.

22 / 5.5 = 4

So we have the point (22, 4). We can do the same for the third value.

33 / 5.5 = 6

So we have the point (33, 6).
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A thin tube stretched across a street counts the number of pairs of wheels that pass over it. A vehicle classified as type A wit
aliina [53]

Answer: All possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

Step-by-step explanation:

Let the number of vehicle of type A be 'x'.

Let the number of vehicle of type B be 'y'.

Since we have given that

Number of axles of type A = 2

Number of axles of type B = 9

After 2 hours, number of counts = 101

so, our equation becomes,

2x+9y=101

So, all possible solutions would be

(1,41), (8,33), (15,25), (22,17), (29,9), (36,1)

7 0
3 years ago
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
Please help Evaluate this equation!!
lozanna [386]

Answer:

1.)-8

2.)-5

3.)4

Step-by-step explanation:

So, the function f(x)=3x-5

The questions say "If you were to substitute these numbers what answer would you get?"

Understanding this,

First question is with x=-1

f(-1)=3(-1)-5

f(-1)=-3-5=-8

_____________________________________

f(0)=3(0)-5

f(0)=0-5=-5

_____________________________________

f(3)=3(3)-5

f(3)=9-5=4

6 0
3 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = (8, 4), v = (9, -9)
Usimov [2.4K]

Answer:

<em>71.6 degrees </em>

Step-by-step explanation:

The formula for calculating the angle between two vectors is expressed as;

u.v = |u||v|cos theta

u.v = (8, 4).(9, -9)

u.v = 8(9)+4(-9)

u.v = 72-36

u.v = 36

|u| = √8²+4²

|u| = √64+16

|u| = √80

|v| = √9²+(-9)²

|v| = √81+81

|v| = √162

36 = √80*√162 cos theta

36 = √12960 cos theta

36 = 113.84 cos theta

cos theta = 36/113.84

cos theta = 36/113.84

cos theta = 0.3162

theta = arccos (0.3162)

<em>theta = 71.6 degrees </em>

<em>Hence the angle between the given vectors is 71.6 degrees </em>

7 0
3 years ago
Read 2 more answers
Trevor is racing his dirt bike at about 30 miles per hour. If he keep a up this rate for an hour and a half, how many miles will
Free_Kalibri [48]

Answer: 45 miles


Step-by-step explanation:

Since Trevor rides 30 mph and does this for 1.5 hours, multiply 30 by 1.5


30 * 1.5 = 45

3 0
3 years ago
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