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mixas84 [53]
3 years ago
6

If you have 98.6% and you get a 85.2% on a test worth 45 points what will be your grade

Mathematics
2 answers:
iris [78.8K]3 years ago
6 0

Answer:

Your Grade Will go up no worries

Step-by-step explanation:

polet [3.4K]3 years ago
6 0
100.79 ? Because the 45 is 2.19% not sure if I’m right sorry
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Use the gcf and the distributive property to find the expression that is equivalent to 35+48
ki77a [65]

Answer:

They all are?

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
To wash a window that is 7 meters off the ground, Roger leans a 9-meter ladder against the side of the building. To reach the wi
iren2701 [21]

Answer:

Roger should place the base of ladder approximately 5.7 m away from base of building to reach the window.

Step-by-step explanation:

Given,

Length of ladder = 9 m

We have to find that how far Roger should place the base of ladder to reach the window.

Solution,

We have drawn the diagram for your reference.

Here the building, the ladder on leaning with land makes a right angle triangle.

Where the length of ladder is 'hypotenuse'.

Height of window from the ground is 'one side of the triangle'.

And distance from base of building to the ladder is 'another side of the triangle'.

So, according to Pythagoras theorem;

"In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of other two sides".

On framing in equation form, we get;

c^2=a^2+b^2

Where 'a' = 1st side of right triangle =7 m

'b' = 2nd side of right triangle

'c' = hypotenuse = 9 m

Now we put the given values and get;

9^2=7^2+b^2\\\\81=49+b^2\\\\b^2=81-49=32\\\\b^2=32

Now taking square root on both side, we get;

\sqrt{b^2} =\sqrt32\\\\b=5.65\ m\approx5.7\ m

Hence Roger should place the base of ladder approximately 5.7 m away from base of building to reach the window.

5 0
3 years ago
Answer a thru d please or give me directions they are all division
xenn [34]
Yup and why would it be hard do it backward as multiplication
5 0
3 years ago
Read 2 more answers
According to these three facts, which statements are true?
Yanka [14]
Q1) We have the following statements:

<span>1. Circle W has center (−3, 0) and radius 8.

We can write this statement as the following equation:

</span>(x-h)^2+(y-k)^2=r^2 \therefore (h,k)=(-3,0) \ and \ r=8 \\ \\ \therefore (x+3)^2+y^2=64
<span>
The graph of this equation is shown in Figure 1.

</span><span>2. Circle V is a translation of circle W, 2 units down.

To do this translation we add two units to the y-coordinate, so:

</span>(x+3)^2+(y+2)^2=64
<span>
The graph is shown in Figure 2.

</span><span>3. Circle V is a dilation of circle W with a scale factor of 2.

To do this dilatation we multiply both the center and the radius by the scale factor of 2, thus:

</span>(x+3)^2+y^2=64 \\ (h,k)=(-3,0) \\ (h_v,k_v)=2\times (-3,0)=(-6,0) \ and \ r_v=8\times2=16 \\ \\ (x+6)^2+y^2=256
<span>
This circle is shown in Figure 3

So let's analyze each statement.

Q1.1) </span><span>The center of circle V is (−5, 0).

This is false. From the statement 2 we know that the new center is (-3, -2) and from 3 the new center is (-6, 0).

Q1.2) 
</span><span>Circle V and circle W are similar.

Since all circles have the same shape even though they may be different sizes, then all circles are similar. Therefore, it is true that circle V and circle W are similar.

Q1.3) 
</span><span>The radius of circle V is 16.

From the statement 3 we can affirm that this is true. By applying the dilatation </span><span>with a scale factor of 2 we find out that the radius of the new circle is in fact equal to 16.

Q1.4) </span>Circle V and circle W have the same center. 

From the statement 2 we know that the new center is (-3, -2) and from 3 the new center is (-6, 0). On the other hand, the center of circle W is (-3, 0). From this, it follows that this statement is false.

Q2) Suppose x is any positive number.

We have two concentric circles as follows:

x\ \textgreater \ 0 \\ \\ Circle \ 1: Center \ (0,0) \ and \ radius \ 2x \\ \\ Circle \ 2: Center \ (0, 0) \ and \ radius \ 10x<span>

Q2.1) Why is circle 1 similar to circle 2?
</span><span>
If we perform a dilatation, that is, a </span>resizing of one circle, centered on the shared center, until both circles overlap, it will be true that the circles will always overlap no matter what size they are, thus, they are similar. In fact, as we said in above all circles are similar. 

Q2.2) 
<span>Circle 2 is a dilation of circle 1 with a scale factor of 0.2.

Suppose that:

</span>x=2
<span>
Then circle 1 is given by the following equation:

</span>x^2+y^2=4
<span>
If we dilate this circle </span><span>with a scale factor of 0.2 then:

</span>Circle \ 1: x^2+y^2=4 \\ Diameter \ 1=2r=2\times 2=4 \\ \\ (h,k)=(0,0) \\ (h_v,k_v)=0.2\times (0,0)=(0,0) \ and \ r_v=0.2\times 2=0.4 \\ \\ Circle \ 2:x^2+y^2=0.16 \\ Diameter \ 2=2\times 0.4=0.8 \\ \\ So: \\ \\ Diameter \ 1 \neq Diameter \2
<span>
So the statement b</span>oth circles have congruent diameters is false.

<span>Q2.3) Circle 2 is a dilation of circle 1 with a scale factor of 5.

We have the same equation for circle 1:

</span>x^2+y^2=4

Circle \ 1: x^2+y^2=4 \\ \\ Dilatation: \\ (h,k)=(0,0) \\ (h_v,k_v)=5\times (0,0)=(0,0) \ and \ r_v=5\times 2=10 \\ \\ Circle \ 2:x^2+y^2=100

Two circles have the same area if they have the same radius. Given that this is no applied to our circles, the statement circle 1 and circle 2 have equal areas is false.

8 0
3 years ago
Can you see if it is this right
Luda [366]
Yep! You’re right! Good job bud! Btw. If you want to be extra you can give out a few more examples
6 0
3 years ago
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