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3 years ago

place an x in the table to show the precent decrease from the first number to the second number. giving brainliest!!

Mathematics

2 answers:

GarryVolchara [31]3 years ago

8 0

Answer:i added a picture, I hope it’s good.

150 - 22% = 117

340 - 25% = 255

450 - 22% = 351

dezoksy [38]3 years ago

4 0

Answer:

150 to 117 is 25%

340 to 255 is 25%

450 to 351 is 22%

Step-by-step explanation:

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Please help! i'm confused! Find the area of a kites with diagonals 9in and 12in.

kolbaska11 [484]

Notice the picture
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3 years ago

Anna can buy 3 sweatshirts for a total of $45.H

Masteriza [31]

Answer: Anna could buy 5 sweatshirts for $60.

Step-by-step explanation:

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weqwewe [10]

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2 years ago

dynasty has a saving account with a starting balance of 250. She decide to add 15 per week write the function notation

tester [92]

Answer:

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Step-by-step explanation:

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3 years ago

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .

$y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$ .

Therefore, the coordinates of point P are also $(1, \ -1)$ .

8 0

2 years ago