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2 years ago

What value digit in 2,417 is ten times as value of the digit in 2,147

Mathematics

1 answer:

3241004551 [841]2 years ago

3 0

Answer:

4 in 2,417

4 in 2,147

Step-by-step explanation:

the place value of 4 in 2,417 is HUNDREDS

while the place value of 4 in 2,147 is TENS

Therefore the place value of 4 in 2,417 is ten times the value of 4 in 2,147

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Melissa and Jane are planting trees. Melissa plants 10 trees in 1 hour. Jane plants 17 tree in 2 hours.

nadezda [96]

Answer:

6 hours

Step-by-step explanation:

Let both of them plan for t hours to plant 111 trees n t hours

Melissa can plants 10 trees in 1 hour

in t hours she can plant 10 t trees

Jane plants 17 tree in 2 hours

in one hour Jane can plant 17/2 trees

in t hours she can plant (17/2)t trees.

Total trees planted by both of them in t hours in terms of t will be

10t + (17/2)t

= (10t*2 + 17t)/2

= (20t + 17t)/2

= 37t/2

but it is given that they planted 111 trees together

so 37t/2 should be equal to 111

37t/2 = 111

=> 37t = 111*2 = 222

=> t = 222/37 = 6.

Thus, it takes 6 hours for both of them to plant 111 trees together.

4 0

2 years ago

Let d(t) be the total number of miles Joanna has cycled, and let t represent the number of hours after stopping for a break duri

Naddik [55]

Answer:

d(4)=?

where

d= distance

t= hours

In this Linear function given by the law d(t) = 12t +20 and the graph below. To find the d(4), means to find out how many miles she cycled in the instant 4 hours.

By this function, after a break , the instant t=0, she had already cycled 20 miles.

Let's calculate

d(4) = 12(4) +20

d(4) =48+20

d(4)= 68 miles

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript

PIT_PIT [208]

Answer:

$f(x)=4\sqrt[3]{16}^{2x}$

Step-by-step explanation:

We believe you're wanting to find a function with an equivalent base of ...

$4\sqrt[3]{4}\approx 6.3496$

The functions you're looking at seem to be ...

$f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x$