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jeka57 [31]
3 years ago
14

Please help I have only got 1 mark for this question and I do not know why I also explained it but it said it was wrong please h

elp, thank you​

Mathematics
1 answer:
Virty [35]3 years ago
3 0
There are 9 apples in the 11th bag
Explanation:
Number of apples x frequency adds up to 87
96-87=9
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Point Bis the midpoint of AC AB=2x and AC=3x+20.<br> What is the value of x?
horsena [70]

Answer:

So x = 20

Step-by-step explanation:

AB + BC = AC sum of a segment is sum of its pieces

AB + AB = AC B is the midpoint, so AB = BC

2x + 2x = 3x + 20 putting in for AB (twice!) and AC

4x = 3x + 20 collect like terms

x = 20 subtract 3x on both sides.

Read more on Brainly.com - brainly.com/question/10838483#readmore

7 0
3 years ago
Help wit this? 10m-8m+2+10
GenaCL600 [577]

Answer:

2m+12

Step-by-step explanation:

add like terms

2m+12

7 0
2 years ago
A manufacturer of cream filled donuts collected data from its automatic filling process, the amount of cream inserted into the d
fenix001 [56]

Answer:

The correct value is t* = 2.797.

Step-by-step explanation:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a  two-tailed confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.797.

4 0
2 years ago
Dos campañas suenas a intervalos de 26 y 78 minutos respectivamente ¿A que hora sonarán juntas otra vez si empieza simultáneamen
faust18 [17]
I need someone to translate for me
4 0
2 years ago
Square roots in trigonometry. I don’t understand please help?
cupoosta [38]

By definitions of the (co)tangent and cosecant function,

3\tan^2x-2=\csc^2x-\cot^2x\iff3\dfrac{\sin^2x}{\cos^2x}-2=\dfrac1{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}

Turn everything into fractions with common denominators:

\dfrac{3\sin^2x-2\cos^2x}{\cos^2x}=\dfrac{1-\cos^2x}{\sin^2x}

Recall that \cos^2x+\sin^2x=1, so we can simplify both sides a bit.

On the left:

\dfrac{3\sin^2x+3\cos^2x-5\cos^2x}{\cos^2x}=\dfrac{3-5\cos^2x}{\cos^2x}

On the right:

\dfrac{1-\cos^2x}{\sin^2x}=\dfrac{\sin^2x}{\sin^2x}=1

(as long as \sin x\neq 0, which happens in the interval 0\le x\le\pi when x=0 or x=\pi)

So we have

\dfrac{3-5\cos^2x}{\cos^2x}=1\implies3-5\cos^2x=\cos^2x

\implies3=6\cos^2x

\implies\cos^2x=\dfrac12

\implies\cos x=\pm\dfrac1{\sqrt2}

\implies x=\dfrac\pi4\text{ or }x=\dfrac{3\pi}4

4 0
2 years ago
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