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3 years ago

12

Subtract 25¢ from $7.42. Include the dollar sign and no spaces in your answer.

1 answer:

yuradex [85]3 years ago

4 0

$7.17 hope this helps

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Mary is inscribing a square in the circle shown, but she is having difficulty remembering the process. Which summary describes t

Ivenika [448]

Answer:

C. Create the perpendicular bisector of FG. Then use the points of intersection of the perpendicular bisector with the circle, along with points F and G, to draw the square.

Step-by-step explanation:

The steps are below:

1. A diameter of the circle is drawn.

2. A perpendicular bisector of the diameter is drawn using the method described in the perpendicular bisector of a segment. This is also a diameter of the circle.

3. The resulting four points on the circle are the vertices of the inscribed square.

4 0

3 years ago

Help please urgent ppleeeaseeeeeeeeeee

jeyben [28]

Answer:

114

Step-by-step explanation:

You will need to multiply L x W and then add them all

8 0

3 years ago

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Dionte needs to make at least $140 over winter break to pay back his parents. If he makes $15 per sidewalk that he shovels for,

lbvjy [14]

15 dollars from 1 sidewalk

140 dollars from ? sidewalks

140/15 =9.333(bar)

Dionte needs to shovel 10 sidewalks

3 0

2 years ago

Guys pls help solve for x

madam [21]

Answer:

x=2

Step-by-step explanation:

x+(2x)^2=18

6 0

3 years ago

Cos(−θ)=√3/3, sinθ<0

Delicious77 [7]

Answer:

$\sin\theta=-\frac{\sqrt{6}}{3}$

Step-by-step explanation:

The given trigonometric equation is $\cos(-\theta)=\frac{\sqrt{3} }{3}$ .

We can either use the Pythagorean identity or the right angle triangle to solve for $\sin\theta$ .

According to the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Recall that, the cosine function is an even function, therefore

$\cos(-\theta)=\cos(\theta)$

$\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}$ .

We substitute this value in to the above Pythagorean identity to get;

$(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1$

$\Rightarrow \frac{3}{9}+\sin^2\theta=1$

$\Rightarrow \sin^2\theta=1-\frac{3}{9}$

$\Rightarrow \sin^2\theta=\frac{6}{9}$

$\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}$

$\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}$

But we were given that,

$\sin\theta\:$ , so we choose the negative value.

$\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}$

The correct answer is B

7 0

3 years ago