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3 years ago

How many times greater is the value of the 5 in 500 than the value of the 5 in 50

Mathematics

1 answer:

densk [106]3 years ago

4 0

It is 10 times bigger. It is easy because you can do 50 x 10 which is 500 and you know it is 10 times bigger.

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Which of the following relationships is a direct variation?

mafiozo [28]

Answer:Well All i know is that the answer is not C I'm sorry I dont know the correct answer but I know for a fact that the answer isn't C

Step-by-step explanation:

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3 years ago

<img src="https://tex.z-dn.net/?f=-8%285d-2%29%3D-35" id="TexFormula1" title="-8(5d-2)=-35" alt="-8(5d-2)=-35" align="absmiddle"

notka56 [123]

-8(5d-2)=-35
-40d+16=-35
-40d= -51
D=-51/-40

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2 years ago

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3 years ago

The string of a flying kite is being held five feet off the ground. It makes a 60 degree angle with one side parallel to the gro

VashaNatasha [74]

Answer:

Maria is right

Height=265 ft

Step-by-step explanation:

In the attached diagram, the length of the kite is BC and the height of the Kite above the ground is CE.

Now, CE=CD+DE

DE=5 ft

Next, we have to determine the length of CD.

In Triangle BCD

$sin\:60^\circ=\frac{|CD|}{300} \\|CD|=300*sin 60\\|CD|\approx260\:ft$

Therefore:

CE=260+5=265 ft

From the above, we see that Maria is right.

Billy erroneously applied the wrong trigonometric ratio (Cosine) which made him get a value of 150ft.

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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