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3 years ago

i need help i hvae done everything but i dont know what i did wrong i did £11.40-£20=£8.60 but it says its wrong how?

Mathematics

2 answers:

Sati [7]3 years ago

8 0

Answer:

£11.43

Step-by-step explanation:

£2.65x2=£5.12

£1.15×3=£3.45

£5.12+£3.45=£8.75

£20.00-£8.75=£11.43

The mistake was in the first line of your answer, you multiplied by 3 instead of 2.

Licemer1 [7]3 years ago

3 0

Answer: It said <u>2</u> chickens, not 3.

Also, when you got your final answer, you have to subtract it from 20, because you want to find the change.

Step-by-step explanation:

This is the mistakes I saw, and I mentioned them above. :)

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Thee questionn is beloww

Phantasy [73]

Given:

$m\angle ABC=60$

To find:

The $m\angle ADC$ .

Solution:

In circle B, $\angle ABC$ is central angle and $\angle ADC$ is inscribed angle from two points A and C.

According to central angle theorem, central angle is always twice of inscribed angle.

$m\angle ABC=2(m\angle ADC)$ [Central angle theorem]

$60=2(m\angle ADC)$

Divide both sides by 2.

$\dfrac{60}{2}=m\angle ADC$

$30=m\angle ADC$

Therefore, $m\angle ADC=30$ .

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2 years ago

Arjun solved the equation 7·m=56 His work is shown. What error did Arjun make?

natulia [17]

He multiplied rather than dividing.

$\mathrm{Divide\:both\:sides\:by\:}7 \ \textgreater \ \frac{7m}{7}=\frac{56}{7} \ \textgreater \ Simplify \ \textgreater \ m = 8$

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2 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

Serga [27]

Answer:

Required solution is $y(x)=1+A\cos x+B\sin x$ where A and B are constants.

Step-by-step explanation:

Given nonhomogenous differential equation is,

$y''(x)+y'(x)=1\hfill (1)$ with $y_1(x)=1$

To find another solution, consider $m=\frac{\partial}{\partial x}$ be such that,

$m^2+1=0\implies m=\pm i$

Hence,

$C.F=A\cos x+B\sin x$ where A and B are constants.

Let $D=\frac{\partial}{\partial x}$

$P.I=\frac{1}{1+D^2}(1)$

$=(1+D^2)^{-1}(1)$

$=(1-D^2+.....)(1)=1$

Hence,

$y(x)=1+A\cos x+B\sin x$ where A and B are constants.

which is required solution.

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3 years ago