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cluponka [151]
3 years ago
8

20 POINTS PLEASEEEE Jessica is bisecting a segment. First, she places the compass on one endpoint and opens it to a width larger

than half of the segment. What is her next step?
Swing an arc on either side of the segment.
Create a point that is not on the segment.
Draw a ray from one endpoint of the segment.
Place the compass on a point that is not on the segment.
Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

9514 1404 393

Answer:

  (a) Swing an arc on either side of the segment.

Step-by-step explanation:

The purpose of placing the compass setting its width is to draw an arc. The next step is ...

  Swing an arc on either side of the segment

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-7(-2a+6)=2a-6(-a+7)
Alenkinab [10]

Answer:

a=0

Step-by-step explanation:

-7(-2a+6)=2a-6(-a+7)

14a-42=2a+6a-42

14a-42=8a-42

6a-42=-42

6a=0

a=0

6 0
3 years ago
Just question 1, please.
ANTONII [103]

Step-by-step explanation:

We can start by writing an equation to describe the data set. For 2D data, one equation form we can use is y=m*x + b, where x is the input and y is the output. Since yuan are put in and dollars are put out, x represents yuan and y represents dollars in this instance.

To find the equation, we can start by plugging in two points, such as (50, 1.25) and (100, 7.5). The slope, or m, is equal to the change in y/change in x, or

(7.5-1.25)/(100-50) = 6.25/50

= 0.125. This means that for each yuan put in, 0.125 dollars come out, according to the formula.

Next, we must find b. Plugging 0.125 in for m and taking a set of points, such as (50, 1.25), we get

1.25 = 50(0.125) + b

1.25 = 6.25 + b

subtract both sides by 6.25 to isolate the b

-5 = b

Therefore, we can write our equation as

y = 0.125 * x - 5

In this equation, there is an exchange rate of 0.125, meaning that for each yuan put in, 0.125 dollars are coming out. The -5 symbolizes that for each transaction, 5 yuan are being lost, which is the fee for exchanging money.

To check if the equation works, we can try it for options such as (200, 20) and (250, 26.25). This does work, so we can move forward with our equation.

For (c), our equation is y = 0.125 * x - 5, with the y representing dollars, x representing yuan, 0.125 representing the exchange rate, and -5 representing the fee for exchaning mone

For (b), because 5 yuan are subtracted from the exchange rate for each transaction, it is fair to assume that there is a fee for exchanging money. There is a fee of 5 yuan per transaction. In dollars, using the exchange rate of 0.125 dollars per yuan, this is equal to 0.625 dollars

For  (a), the exchange rate is 0.125 dollars per yuan. We know this because for each yuan put in, 0.125 dollars are coming out after taking into account the fee

6 0
3 years ago
Consider the geometric sequence: 3, 15, 75, 375, 1875 ... 1. What is the constant ratio, r, for this sequence? 2. Use r to write
zalisa [80]
R= 5

a_{n} =3* 5^{n-1}


7 0
3 years ago
Read 2 more answers
2x ^2 −5=3x
expeople1 [14]

Answer:a

Step-by-step explanation:

3 0
3 years ago
Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an
Natasha_Volkova [10]

Answer:

The volume of the solid is 243\sqrt{2} \ \pi

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0  ≤ ρ²   ≤   81

0  ≤ ρ   ≤  9

The intersection that takes place in the sphere and the cone is:

x^2 +y^2 ( \sqrt{x^2 +y^2 })^2  = 81

2(x^2 + y^2) =81

x^2 +y^2 = \dfrac{81}{2}

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

z = \sqrt{x^2+y^2}

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; \dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}

Thus, volume: V  = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4}  \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho   ^2 \ sin \phi \ d\rho \   d \theta \  d \phi

V  = \int \limits^{\pi/2}_{\pi/4} \ sin \phi  \ d \phi  \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho   ^2 d\rho

V = \bigg [-cos \phi  \bigg]^{\pi/2}_{\pi/4}  \bigg [\theta  \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3}  \bigg ]^{9}_{0}

V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]

V = 243\sqrt{2} \ \pi

4 0
3 years ago
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