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erma4kov [3.2K]
3 years ago
5

-6t-7=17 what does t equal

Mathematics
1 answer:
Grace [21]3 years ago
7 0

Answer:

t=-4

Step-by-step explanation:

I know you don't care about any explanation, and just want the answer, but to get this, you can add 7 to both sides. Then you get -6t=24. Divide both sides by -6 and you get -4

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The right isosceles triangle shown is rotated about line k with the base forming perpendicular to k. The perimeter of the triang
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Answer:

The new shape will be cone with a radius of 17 units, tilt height 24 and units of height k.

Step-by-step explanation:

The image related to the exercise is necessary, to be able to solve therefore the attached one.

We have the following information:

Perimeter of the triangle = 58 units

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We have that the other two sides are equal and have x units, therefore we have the following:

x + x + 24 = 58

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4 0
3 years ago
Someone please be awesome and help me please :(
solong [7]

Answer:

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

Step-by-step explanation:

x^2+\frac{b}{a}x+\frac{c}{a}=0

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared.  Whatever you add in, you must take out.

x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

Now we are read to write that one part (the first three terms together) as a square:

(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

I don't see this but what happens if we find a common denominator for those 2 terms after the square.  (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0

They put it in ( )

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

I'm going to go ahead and combine those fractions now:

(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}

x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}

Now subtract b/(2a) on both sides:

x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

Combine the fractions (they have the same denominator):

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

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