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3 years ago

9

Two shops sell baked beans in different sized tins. Calculate the price per 2.5 kg for shop B. Give your answer in pence. Write

the shop that is better value in the comments.

2 answers:

RUDIKE [14]3 years ago

8 0

Answer:

Step-by-step explanation:

846p

defon3 years ago

6 0

Answer:

You have not given all information

Step-by-step explanation:

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Karra is at the dollar store buying bags of chips for her sleepover party. For every 5 bags of chips she buys, she pays $6. What

Alexus [3.1K]

Answer:

$1.20

Step-by-step explanation:

Given that:

Price paid to buy 5 bags of chips = $6

To find:

Price per bag of chips in two digits after the decimal.

Solution:

To find the price of each bag of chips, we need to divide the price of 5 bags of chips with 5.

i.e. we need to divide $6 with 5, then we will get the price of each bag of chips.

$\text{Price of each bag of chips}=\dfrac{\text{Price of 5 bags of chips}}{5}\\\Rightarrow \dfrac{\$6}{5} = \bold{\$1.2}$

Therefore, the answer is:

Price of each bag of chips is <em>$1.2.</em>

4 0

3 years ago

Bridgette made a graph of her neighborhood on a coordinate grid. The points A(-2, 3), B(-2, -2), C(2, -2), and D(2, 3) represent

vodka [1.7K]

Answer:

$\boxed{\text{18 blocks}}$

Step-by-step explanation:

Distance travelled on graph = AB + BC + CD + DA

AB = CD = (3 +2) = 5

BC = AD = 2 –(-2) = 4

Distance on graph = 2 × 5 + 2 × 4 = 10 + 8 = 18 spaces = 18 blocks

Bridgette jogs $\boxed{\textbf{18 blocks}}$ each day.

8 0

3 years ago

Please please help!!

Olenka [21]

Answer:

The formula T= 10d +20

A) what does each term on the right side of the equation represent?

10d⇒ 10 degrees increase per 1 km and 20 deg surface temperature

B) Estimate the depth where the temperature is 60 degrees C.

60=10d+20
10d=40
d=4 km

C) What is the approximate temperature at a depth of 4km?

T=10*4+20
T=60 deg

4 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

What is the positive solution to this equation 4x^2+12x=135

Stels [109]

Answer:

answer for the equation is 6.75

5 0

3 years ago