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Sindrei [870]
3 years ago
5

Abi and Ali shared £35 in the ratio 1 : 6. Work out how much Ali received.

Mathematics
2 answers:
Ray Of Light [21]3 years ago
6 0

Answer:

£30

Step-by-step explanation:

6+1=7

35/7=5

5 X 6 = 30

5 X 1 = 5

1 : 6 = 5 : 30

so Ali has 30

baherus [9]3 years ago
4 0

Answer:

£30

Step-by-step explanation:

To tackle these 'sharing' ratios problems, first find the total amount parts in the ratio.

1 : 6 = 1 + 6 = 7 parts

Now divide the total amount of parts by the total amount of money shared

35 / 7 = 5

Now multiply the answer by the value that is appropriate.

In this case, they want you to work out how much Ali recieved.

Ali's value is 6

6 × 5 = £30

hope this helps :)

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Ashley took out a $1500 loan and promises to pay it back with 5% interest after 18 months. How much interest would she have to p
babymother [125]

Answer:

75.00

Step-by-step explanation:

1500*0.05

because 5% is her interest you convert 5% into a decimal = 0.05

then you multiply 1500 by 0.05 and you get 75.00

5 0
3 years ago
Solve for n. then find the value of m for each value of n<br> 8n=-3m+1
Eva8 [605]
Basically, solve for n, then find aanother tha is solve for m


8n=-3m+1
divide both sides by 8
n=-3/8m+1/8

and find m
8n=-3m+1
minus 1 both sides
8n-1=-3m
divide both sides by -3
-8/3n+1/3=m
m=-8/3n+1/3

6 0
3 years ago
A triangle has two side of length 3 and 7. The third side of this triangle must not be shorter than (blank) or longer than (blan
Marina86 [1]

Answer:

The sum of two side lengths of a triangle must be greater than the length of the third.

So, if two side lengths are 3 and 7, the length of the third side must be less than 10 units.

However, it cannot be shorter than 4 units, as 3 + 4 = 7.

So your answer is: The third side of this triangle must not be shorter than 4 or longer than 10

Let me know if this helps!

5 0
3 years ago
Express the given integral as the limit of a riemann sum but do not evaluate: the integral from 0 to 3 of the quantity x cubed m
WARRIOR [948]
We will use the right Riemann sum. We can break this integral in two parts.
\int_{0}^{3} (x^3-6x) dx=\int_{0}^{3} x^3 dx-6\int_{0}^{3} x dx
We take the interval and we divide it n times:
\Delta x=\frac{b-a}{n}=\frac{3}{n}
The area of the i-th rectangle in the right Riemann sum is:
A_i=\Delta xf(a+i\Delta x)=\Delta x f(i\Delta x)
For the first part of our integral we have:
A_i=\Delta x(i\Delta x)^3=(\Delta x)^4 i^3
For the second part we have:
A_i=-6\Delta x(i\Delta x)=-6(\Delta x)^2i
We can now put it all together:
\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\&#10;\sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]
We can also write n-th partial sum:
S_n=(\frac{3}{n})^4\cdot \frac{(n^2+n)^2}{4} -6(\frac{3}{n})^2\cdot \frac{n^2+n}{2}

4 0
3 years ago
Does anyone know how to do this <br><br>csc x ( sin x + cos x) = 1 + cot x ​
Dima020 [189]

9514 1404 393

Explanation:

Use the identities ...

  csc(x) = 1/sin(x)

  cot(x) = cos(x)/sin(x)

__

  \csc(x)(\sin(x)+\cos(x))=1+\cot(x)\\\\\dfrac{1}{\sin(x)}(\sin(x)+\cos(x))=1+\cot(x)\\\\\dfrac{\sin(x)}{\sin(x)}+\dfrac{\cos(x)}{\sin(x)}=1+\cot(x)\\\\1+\cot(x)=1+\cot(x)

8 0
3 years ago
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