Please help!
Question:
Divide £40 in the ratio 3 : 7

Answer:

5

Step-by-step explanation:

8-1-(18-2)÷8

PEMDAS says parentheses first

8-1-16÷8

Then divide

8-1-2

Then subtract from left to right

7-2

5

Answer:

[tex]A=2044[/tex]

Step-by-step explanation:

Note that [tex]x\in\mathbb{W}[/tex] denotes that [tex]x[/tex] is a whole number.

By definition, consecutive numbers follow each other when we count up (e.g. 1, 2, 3).

Let's consider our conditions:

Since B is a multiple of 5, the ones digit of B must be either 0 or 5. However, notice that the number before it, A, needs to be a multiple of 4. The ones digit of a number preceding a ones digit of 0 is 9. There are no multiples of 4 that have a ones digit of 9 and therefore the ones digit of B must be 5.

Because of this, we've identified that the ones digit of A, B, and C must be 4, 5, and 6 respectively.

We can continue making progress by trying to identify the smallest possible whole number greater than 2,000 with a units digit of 6 that is divisible by 6. Notice that:

[tex]2000=2\mod6[/tex]

Therefore, [tex]2000-2=1998[/tex] must be divisible by 6. To achieve a units digit of 6, we need to add a number with a units digit of 8 to 1,998 (since 8+8 has a units digit of 6).

The smallest multiple of 6 that has a units digit of 8 is 18. Check to see if this works:

[tex]C=1998+18=2016[/tex]

Following the conditions given in the problem, the following must be true:

[tex]A\in \mathbb{W},\\B\in \mathbb{W},\\C\in \mathbb{W},\\A+1=B=C-1,\\A=0\mod 4,\\B=0\mod 5,\\C=0\mod 6,[/tex]

For [tex]C=2016[/tex], we have [tex]B=2015[/tex] and [tex]A=2014[/tex]:

[tex]A\in \mathbb{W},\checkmark\\B\in \mathbb{W},\checkmark\\C\in \mathbb{W},\checkmark\\A+1=B=C-1,\checkmark\\A=2014\neq 0\mod 6, \times\\B=2015=0\mod 5,\checkmark\\C=2016=0\mod 6\checkmark\\[/tex]

Not all conditions are met, hence this does not work. The next multiple of 6 that has a units digit of 8 is 48. Adding 48 to 1,998, we get [tex]C=1998+48=2046[/tex].

For [tex]C=2046[/tex], we have [tex]B=2045[/tex] and [tex]A=2044[/tex]. Checking to see if this works:

[tex]A\in \mathbb{W},\checkmark\\B\in \mathbb{W},\checkmark\\C\in \mathbb{W},\checkmark\\A+1=B=C-1,\checkmark\\A=2044=0\mod 4,\checkmark\\B=2045=0\mod 5,\checkmark\\C=2046=0\mod 6\checkmark[/tex]

All conditions are met and therefore our answer is [tex]\boxed{2,044}[/tex]

Answer: Choice A (both are smiley faces)

This is because the reflection doesn't flip the faces upside down or anything (instead it does a left-right swap in a way). This is why both faces are smiley faces.

Answer:

360cm³

Step-by-step explanation:

Volume of a triangular prism = Base area * Height of prism

Height of prism = 15cm

Base area = 1/2 * 6 * 8

Base area = 24cm²

Volume of the prism = 15 * 24

Volume of the prism = 360cm³

​

Answer:

3/(10-x) = 5/(10-x+x)

3/(10-x) = 1/2

6 = 10-x

x = 4

Hope this helps!

Answer:

37

Step-by-step explanation:

360 -(110+110+66) = 74

74/2 = 37

Answer:

lower limit = 5

upper limit = 10

class mark = lower limit + upper limit ÷ 2 = 7.5

class size = upper limit - lower limit

= 10-5 = 5

Step-by-step explanation:

I hope this help you have an great day and safe health

Answer:

u should use a ruler and multiplied

Step-by-step explanation:

I dont know the step by step sorry

Answer:

m is slope and y is the intercept

Answer: 5x and -4x

Step-by-step explanation:

= 5x + (-4x)

= 5x -4x

= +x

= 5x × (-4x)

= -20x²

please click thanks and mark brainliest if you like :)

Answer: x = 2 and y = -4

x + 2y = -6

x = -6-2y

Putting this in value of x in

6x + 2y = 4

6(-6-2y) + 2y = 4

-36-12y+2y = 4

-10y = 4+36

y = 40/(-10)

y = -4

Now putting this value of y in

x + 2y = -6

x + 2(-4) = -6

x -8 = -6

x = -6+8

x = 2

Therefore x = 2 and y = -4

If we put these values we can check this

x + 2y = -6

2 + 2(-4) = -6

2 -8 = -6

-6 = -6

please click thanks and mark brainliest if you like :)

Answer:

Step-by-step explanation:

6. Yes: these two angles are opposite 2 equal line.

7. No: We don't know that AC bisects <JAD

8. No: We don't know that JCA or DCA are right angles.

9. Yes. Both angles have AC in common. JD is a straight line which has 180 degrees. AC intersects a JD and creates 2 angles. We know their total but not their equality.

10. Yes. They are given as equal.

Not everything you see is true just because it looks that way.

Answer:

[tex] 3x + y = 2 [/tex]

Step-by-step explanation:

[tex] y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1) [/tex]

[tex] y - 5 = \dfrac{-7 - 5}{3 - (-1)}(x - (-1)) [/tex]

[tex] y - 5 = \dfrac{-12}{4}(x + 1) [/tex]

[tex] y - 5 = -3(x + 1) [/tex]

[tex] y - 5 = -3x - 3 [/tex]

[tex] 3x + y = 2 [/tex]

Answer:

Step-by-step explanation:

The dimensions as a pair, appear twice for each combination.

SA = 2* 16 * 8 + 2 * 16 * 6 + 2*8* 6

That's because when you look at the figure, there are 2 places each face is positioned.

SA = 256 + 192 + 96

SA = 544

Answer:

True.

Step-by-step explanation:

[tex]\frac{(10)-2}{4}=2\\\\\frac{8}{4}=2\\\\2=2[/tex]

Answer:

k = 11

Step-by-step explanation:

What the factor theorem is telling us is that if x + 1 is a factor of 2x - kx²-8x + 5 then when x = - 1 is put into the equation, the result will be 0. That's because x +1 = 0 when x  = - 1

f(-1) = 2(-1) - k(-1)^2 - 8(-1) + 5

f(-1) = -2 - k(1) + 8 + 5 = 0

-k*(1) + 11 = 0

-k = -11

k = 11

Answer:

here I'm confused that whether the selling price is Rs.9,200 or Rs.9.200

any ways you can take the help of below procedure

Step-by-step explanation:

here,

let marked price be X

discount(d)=20%

VAT=15%

SP with VAT= 9200

now,

SP without VAT= SP - VAT of SP

= 9200-15/100 ×9200

= 9200- 1380

=Rs. 1380

again,

SP= MP - D of MP

or,7820= x- 20/100 × x

or, 7820× 100 = 100x- 20x

or,782000=80x

or, 782000/80=X

or, X= 9775.

hence MP is Rs 9775

Answer:

Slope: -5/4

Step-by-step explanation:

Slope formula: [tex]\frac{y^2-y^1}{x^2-x^1}[/tex]

Plug in:

[tex]\frac{-4-6}{5-(-3)}[/tex]

Solve:

[tex]\frac{-4-6}{5-(-3)}[/tex]

-4 - 6 = -10

5-(-3) = 8

-10              5

-----   =   -  -----

8               4

The answer is -5/4

Hope this helped.

Answer:

Step-by-step explanation:

Find the slope of the line

(x₁ , y₁) = (6 , 8)   & (x₂ , y₂) = (0 , 5)

[tex]Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{5-8}{0-6}\\\\=\frac{-3}{-6}\\\\=\frac{1}{2}[/tex]

m = 1/2  ;(x₁ , y₁) = (6 , 8)

y - y₁ = m(x - x₁)

[tex]y - 8 = \frac{1}{2}(x - 6)\\\\y - 8 =\frac{1}{2}x -\frac{1}{2}*6\\\\y - 8 =\frac{1}{2}x- 3\\\\y = \frac{1}{2}x - 3 + 8\\\\y = \frac{1}{2}x + 5[/tex]

Answer:

a + b = 5.30

a + b = 7

No

Step-by-step explanation:

Expressing the information as system of linear equation :

Let apples = a, oranges = b

If $5.30 is charged for one apple and one orange, then we have ;

a + b = 5.30 - - - (1)

If $14 is charged for 2 apples and 2 oranges, then we have ;

2a + 2b = 14 - - - - (2)

a + b = 7

Since both equations gives varying combined cost for equal amount of the fruit, then a unique cost cannot be obtained for each fruit from the systems of equation using simultaneous equation process.

From (1)

a = 5.30 - b

Put a = 5.30 - b in (2)

2(5.30 - b) + 2b = 14

10.6 - 2b + 2b = 14

10.6 = 14 - - - - - (variables cancels out).

Answer:No the System has no solution

Step-by-step explanation:

Answer:

Not enough infomation, you need to provide #12

Step-by-step explanation:

Answer:

what can i help u with

Step-by-step explanation:

I really can't help u with that sorry i am bad at math

Hi there!

[tex]\large\boxed{102^o}[/tex]

Using the triangle exterior angle theorem, we know that:

∠4 = the other two angles of the triangle combined (both are given)

Thus:

∠4 = 51 + 51 = 102°

Angle 4 is an outside angle.

An outside angle of a triangle is equal to the sum of the two opposite inside angles.

The two opposite inside angles to angle 4 are shown as 51 and 51

Angle 4 = 51 + 51

Angle 4 = 102 degrees.

I assume you're supposed to find the limit as n approaches infinity, not x.

You have

[tex]\displaystyle \lim_{n\to\infty}\frac{4^n-5.3^n+1}{2.4^n+2} = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-\left(\dfrac{5.3}{5.3}\right)^n+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}} \\\\ = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-1+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}}[/tex]

For |x| < 1, we have lim |x|ⁿ = 0 as n goes to infinity. Then each exponential term converges to 0, which leaves us with -1/0. This means the limit is negative infinity.

On the other hand, perhaps you meant to write

[tex]\displaystyle \lim_{n\to\infty}\frac{4^n-5\times3^n+1}{2\times4^n+2}[/tex]

The same algebraic manipulation gives us

[tex]\displaystyle\lim_{n\to\infty}\frac{\left(\dfrac44\right)^n-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2\left(\dfrac44\right)^n+\dfrac2{4^n}} = \lim_{n\to\infty}\frac{1-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2+\dfrac2{4^n}}[/tex]

Again the exponential terms converge to 0, but this time we're left with the limit 1/2.

Answer:

Step-by-step explanation:

If the account grows steadily, this is a linear representation, where the constant, steady growth translates to the slope of a line.

Answer:

V=l×w×h

so the answer is 66km

11×2×3=66

Answer:

x = 9

Step-by-step explanation:

When doing these types of problems, the main goal is to get the variable by itself. Its kind of hard to explain with words so I will just use numbers.

15x + 14 = 149

(Subtract 14 on both sides. Remember, when you do something (Like subtracting, adding, etc.) on one side you have to do it to the other.)

15x + 14 = 149

       -14 =  -14

---------------------------

15x = 135

(Divide 15 on both sides. Since we need to use inverse operation to cancel the 15 out, we would divide 15 on both sides.)

15x/15 = 135/15

x = 9

Hope this helps :)

Answer:

Step-by-step explanation:

[tex]\sf 15x + 14 = 149[/tex]

[tex]\sf 15x+14-14=149-14[/tex]

[tex]\sf 15x=135[/tex]

[tex]\sf \cfrac{15x}{15}=\cfrac{135}{15}[/tex]

[tex]\sf x=9[/tex]

The derivative of a function f(x) is defined as

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]

For f(x) = 3x ² + 4, we have

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{(3(x+h)^2+4) - (3x^2+4)}h[/tex]

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{(3(x^2+2xh+h^2) - 3x^2}h[/tex]

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{6xh+3h^2}h[/tex]

[tex]f'(x)=\displaystyle\lim_{h\to0}(6x+3h) = \boxed{6x}[/tex]