All categories

2 years ago

Jack spends 1 1/2 times as long on his homework as Jill.

Mathematics

2 answers:

Mashcka [7]2 years ago

6 0

Answer:

12 3/4

Step-by-step explanation:

8.5 times 1 is 8.5

and 8.5 times .5 is 4.25

so 8.5 plus 4.25 is 12.75 or 12 3/4

Minchanka [31]2 years ago

4 0

Answer:

12 3/4 hours

Step-by-step explanation:

Jack spends 1 1/2 times as long on his homework as Jill .

Jill spends 8 1/2 in doing his home work .

Time spent by jack in doing his homework is jacks time multiplied by Jill’s

That’s

1 1/2 x 8 1/2

Change to improper fractions.

We have,

3/2 x 17/2

Multiply through

51/4 = 12 3/4

Jack spends 12 3/4 hours in doing his homework

You might be interested in

"The graph of F(x), shown below, resembles the graph of G(x) = x2, but it has been changed somewhat. Which of the following coul

SpyIntel [72]

G(x)=x²
The graph has moved to the right 4 units, therefore the new graph will be:
H(x)=(x-4)²

It has also move 4 units up, therefore the new graph will be:
F(x)=(x-4)²+4

Answer:
F(x)=(x-4)²+4

5 0

3 years ago

Read 2 more answers

I really need help with this ASAP

Alexandra [31]

Answer:

No. of boxes = 10

Oranges per box = 56

Total no. of oranges = 56*10=560

No. of bad oranges = 560/40= 14

(i) Probability of bad orange = 14/560 = 1/40

(ii) no of oranges expected to be bad = 14 ( found above)

Hope it helps.............. :)

7 0

3 years ago

I need answer ASAP I have 30 minutes and 3 more questions to answer. SHOW YOUR WORK PLEASE.

Gnom [1K]

Answer:

6 weeks

Step-by-step explanation:

6*5=30

100+30=130

20*6=120

120+10=130

130=130

5 0

2 years ago

Julio bought a total of 20 medium and large pumpkins. If he spent $53 and bought 6 more large pumpkins as medium pumpkins, how m

Tatiana [17]

20 medium and large pumpkins

m ... number of medium pumpkins
l ... number of large pumpkins

20 = m + l
20 = m + (m + 6)
20 = 2 * m + 6
2 * m = 20 - 6
2 * m = 14
m = 7 medium pumpkins

l = 20 - 7 = 13 large pumpkins

Result: Julio bought 13 large pumpkins.

5 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago