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CaHeK987 [17]
2 years ago
10

Make y the subject z+2 = 2/1-y

Mathematics
1 answer:
s344n2d4d5 [400]2 years ago
3 0
Y= subject or y1/2=2
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John wants to find the center of a wall so he can hang a picture. He measures the wall and determines it is 65.25" wide. 65.25"
antiseptic1488 [7]

Option B is correct.

John wants to find the center of a wall so he can hang a picture. He measures the wall and determines it is 65.25" wide.

Here, 65.25" is Quantitative, continuous

There are two types of quantitative data or numeric data: continuous and discrete.

As a general rule, counts are discrete and measurements are continuous. A continuous data can be recorded at many different points (length, size, width, time, temperature, etc.)

So, option B is the answer.


3 0
3 years ago
Please help me with this question I would really appreciate it!!!
shepuryov [24]

Answer:

0, 1.32

Step-by-step explanation:

16cos(t)*sin(t)=4sin(t)

cos(t)=1/4 or sin(t)=0

t=1.32 or 0

8 0
3 years ago
The smiths have two children. The sum of their ages is 23. The produce of their ages is 132. How old are the children?
kolezko [41]

For this case we propose a system of equations:

x: Let the variable representing the age of the first child of the Smiths

y: Let the variable representing the age of the second child of the Smiths

According to the data of the statement we have to:

x + y = 23\\x * y = 132

From the first equation we have to:

x = 23-y

We substitute in the second equation:

(23-y) * y = 132\\23y-y ^ 2 = 132\\y ^ 2-23y + 132 = 0

We find the solutions by factoring:

We look for two numbers that, when multiplied, result in 132 and when added, result in 23. These numbers are 11 and 12.

Thus, we have that the factorized equation is:

(y-11) (y-12) = 0

Thus, the solutions are:y_ {1} = 11\\y_ {2} = 12

So, we can take any of the solutions:

With y = 11

Thenx = 23-11 = 12

Therefore, the ages of the children are 11 and 12 respectively.

Answer:

 The ages of the children are 11 and 12 respectively.

6 0
3 years ago
Which wind belt starts at the horse latitudes and moves toward the poles?
kvasek [131]

Answer:

The prevailing westerlies.

Step-by-step explanation:

The prevailing westerlies are found between 30 (horse latitudes) and 60 degrees latitude. They come from the high-pressure areas in the horse latitudes and go towards the poles.

4 0
4 years ago
Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
Ede4ka [16]

Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

5 0
3 years ago
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