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Verizon [17]
3 years ago
5

Given that 2 years ago, Mia's age was half of the age she will be in 7 years. How old is she now?

Mathematics
1 answer:
QveST [7]3 years ago
3 0

Answer:

Step-by-step explanation:45779756

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A watercolor painting is 19 inches long by 11 inches wide. Ramon makes a border around the watercolor painting by making a mat t
Alexxx [7]

Answer:

14

Step-by-step explanation:

3 0
3 years ago
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At a 3​-mile ​cross-country race, an athlete runs 2 miles at 6 mph and 1 mile at 12 mph. What is the​ athlete's average​ speed?
Tom [10]

Answer: the average speed is 7.5 mph or at least im pretty sure

Step-by-step explanation:

This is because if you make the 2 miles at 6 mph to 1 mile at 3 mph then you get 1 mile at 3mph and 1 mile at 12 mph. now that you have the unit rates of these you can add them. 1 + 1 = 2 miles and 3 + 12 = 15 mph now divide this by two and you get 7.5 mph per mile.

I hope this helped, if it did could you give me a good rating? lol

8 0
3 years ago
Help please I have like 20 minutes to complete this
Nataliya [291]

Answer:     b

Step-by-step explanation:

good luck... ur profile pick is hot

3 0
3 years ago
Read 2 more answers
Find the diameter and radius of a circle with a circumference of 65.98 Please help ​
viva [34]

Answer:

21 and 10.5 respectively

Step-by-step explanation:

Remember circumference of a circle is given as;

C= 2×π×r; r is raduis

r = C / 2×π

=65.98/(2×3.142)= 10.50

D= 2× r = 2× 10.50= 21.0( D represent diameter)

Note π = 3.142 a known constant

6 0
3 years ago
Kaliska is jumping rope. The vertical height of the center of her rope off the ground R(t) (in cm) as a function of time t (in s
xz_007 [3.2K]

Answer:

R (t) = 60 - 60 cos (6t)

Step-by-step explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

R(0) = 0

0 = acos (0) + d

a + d = 0 ----- (1)

After \dfrac{\pi}{12} seconds it reaches a height of 60 cm from the ground.

i.e

R ( \dfrac{\pi}{12}) = 60

60 = acos (\dfrac{b \pi}{12}) +d --- (2)

Recall from the question that:

At t = 0, R(0) = 0 which is the minimum

as such it is only  when a is  negative can acos (bt ) + d can get to minimum at t= 0

Similarly; 60 × 2 = maximum

R'(t) = -ab sin (bt) =0

bt = k π

here;

k  is the integer

making t the subject of the formula, we have:

t = \dfrac{k \pi}{b}

replacing the derived equation of k into R(t) = acos (bt) + d

R (\dfrac{k \pi}{b}) = d+a cos (k \pi) = \left \{ {{a+d  \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.

Since we known a < 0 (negative)

then d-a will be maximum

d-a = 60  × 2

d-a = 120 ----- (3)

Relating to equation (1) and (3)

a = -60 and d = 60

∴ R(t) = 60 - 60 cos (bt)

Similarly;

For R ( \dfrac{\pi}{12})

R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60

where ;

cos (\dfrac{\pi b}{12}) =0

Then b = 6

∴

R (t) = 60 - 60 cos (6t)

7 0
3 years ago
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