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3 years ago

What is the slope of y = -1+ 3.x?

Mathematics

1 answer:

zysi [14]3 years ago

5 0

Answer:

The slope is 3.

Step-by-step explanation:

y=-1+3x

y=3x-1

y=mx+b where m=slope and b=y-intercept

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In a choir there are 27 men and 60 women.

guapka [62]

Answer:

b) A person is chosen at random from the choir.

Work out the probability that the person.

Work out the probability that the person does not wear glasses

8 0

2 years ago

What is the lateral area of the cone to the nearest whole number? The figure is not drawn to

Sveta_85 [38]

Answer:

The lateral surface area of the cone is $25120\ m^2$ .

Step-by-step explanation:

We have.

Height of a cone is 60 m

Diameter of a cone is 10 m

Radius, r = 80 m

The lateral surface area of a cone is given by :

$A=\pi rl$

l is slant height of the cone,

$l=\sqrt{60^2+80^2} =100\ m$

Plugging all values in above formula,

$A=3.14\times 80\times 100\\\\A=25120\ m^2$

So, the lateral surface area of the cone is $25120\ m^2$ .

6 0

3 years ago

Why is ten a good number ?

timurjin [86]

10 is a good number because is what adds to 100 and the number 100 is lucky. This is what i think.

8 0

2 years ago

Jacob sells his motorbike for £112 making a loss of 12% on its original price. Work out its original price

liq [111]

Answer:

125,44

Step-by-step explanation:

Loss = 112÷100*12

=13,44

Therefore : 112 + 13%

=112 + 13,44

=125,44

4 0

2 years ago

How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title

kotegsom [21]

Let $t=\sqrt y$ , so that $t^2=y$ , $t^4=y^2$ , and $\mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}$ . Then

$\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}$

Now let $y=\sqrt2\tan z$ , so that $\mathrm dy=\sqrt2\sec^2z\,\mathrm dz$ . Then

$\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C$

Transform back to $y$ to get

$\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C$

and again to get back a result in terms of $t$ .

$\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C$

3 0

3 years ago