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3 years ago

Joe has read 30% of a book. He has 49 more pages to finish. How many pages are there in the book?

Mathematics

1 answer:

sasho [114]3 years ago

7 0

Answer:

70 pages

Step-by-step explanation:

100%-30%=70%

70% of the book is 49 pages.

x is the total number of pages.

0.7x=49

7x=490

x=70

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if triangle ABC is an isosceles triangle and triangle DBE is an equilateral triangle, find each missing measure

BabaBlast [244]

Picture?? This question is incomplete just get a picture then I can answer it

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3 years ago

There are 5 roads from Allen to baker, 7 roads from baker to Carlson, and 4 roads from Carlson to dodge. How many different rout

Basile [38]

Answer:

140 routes

Total Number of roads from allen to dodge through baker and Carlson is 140 routes.

Step-by-step explanation:

Given;

Number of roads from Allen to baker = 5

Number of roads from baker to Carlson = 7

Number of roads from Carlson to dodge = 4

Total Number of routes from allen to dodge through baker and Carlson is;

N = 5×7×4

N = 140 routes

6 0

2 years ago

Read 2 more answers

PLEASE HELP 7TH GRADE MATH

grigory [225]

Answer:

B. 224

Step-by-step explanation: 42/60 means that 70 % of people ride the school bus. So 70 percent of 320 is 224. Hope this helps

7 0

2 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

How do i solve this inequality? Number 46

myrzilka [38]

6(x-2.5) ≥ 8-6(3.5+x)

A number is before a parenthesis. The first step is to multiply.

6(x-2.5) = 6x -15

8-6(3.5+x)
2(3.5+x)= 7+2x

6x-15≥7+2x

Now solve

6x-15≥7+2x
-2x -2x

(-2x+2x)= They cancel out

(6x-2x)= 4x

4x-15≥7
+15 +15

4x≥22
/4 /4

You get= x≥5.5

7 0

3 years ago