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ss7ja [257]
2 years ago
11

Please asap maths homework

Mathematics
1 answer:
Nutka1998 [239]2 years ago
6 0

Answer:

77 red marbles

Step-by-step explanation:

the bag has R and B marbles

P(B)=1/12 = number of blue /total number of marbles

so 1/12=?/84

in the bag must be 84/12=7 marbles that are blue

84-7=77 red marbles

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Sides:6.7,9.3,9.5

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In the diagram below, BD is parallel to XY. what is the value of y?
melisa1 [442]

I can't see the diagram sorry.

Step-by-step explanation:

Is there supposed to be a picture attached?

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2 years ago
There are 40 female performers in a dance recital. The ratio of men to women is 3:5. How many men are in the dance recital?
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2 years ago
In ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. Find the measure of ∠J to the nearest degree.
nydimaria [60]

We have been given that in ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. We are asked to find the measure of angle J to nearest degree.

First of all, we will draw a triangle as shown in the attachment.

We can see from our attachment that side KL is opposite side to angle J and side JK is hypotenuse of right triangle.

We know that sine relates opposite side of right triangle to hypotenuse.

\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}

\text{sin}(\angle J)=\frac{22}{54}

Using inverse sine or arcsin, we will get:

\angle J=\text{sin}^{-1}(\frac{22}{54})

\angle J=24.042075905756^{\circ}

Upon rounding to nearest degree, we will get:

\angle J\approx 24^{\circ}

Therefore, the measure of angle J is approximately 24 degrees.

8 0
3 years ago
Find limit as x approaches 2 from the left of the quotient of the absolute value of the quantity x minus 2, and the quantity x m
Molodets [167]

Answer:

\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

Step-by-step explanation:

We want to find \lim_{x \to 2^-} \frac{|x-2|}{x-2}.

By definition:

|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1).

The limit of a constant is the constant.

\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1.

8 0
2 years ago
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