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a + b = 26.

To write a number or decimal in scientific notation, there has to be a number that isn't zero in the ones place. So, 0.00175 would be written as 1.75 times 10³.

Hope this helped!!!

In short, the answer is 33 numbers.

I used a 100s table and eliminated all multiples of 2 and 3 and then counted the remaining numbers.

I'm sure there are other faster ways to do this but this was the first way that came to mind.

Hope it helps!

Let an angle BAD be  β       β = 7º

1/       AD = cosβ * 3.7

2/      BD = sinβ * 3.7

3/      A_(ΔABD) = (AD * BD) / 2

4/      A_(ΔADC) = A(ΔABD) + 22.4

5/     CD = (2 * A_(ΔADC)) / AD

6/     b = sqrt( AD² + CD² )   

In a quadrilateral ABCD , AB || DC . If angle A : angle D = 2:3 and angle B : angle C = 7:8 , find the measure of each angle​

Hello Guest!

\(\alpha :(180^{\circ} -\alpha )=2:3\\ 3\alpha =360^{\circ} -2\alpha\\ 5\alpha=360^{\circ} \)

\(\alpha =72^{\circ}\)

\(\beta:(180^{\circ} -\beta)=7:8\\ 8\beta=1440^{\circ} -7\beta\\ 15\beta=1440^{\circ} \)

\(\beta=96^{\circ} \)

\(\gamma=180^{\circ} -\beta=180^{\circ} -96^{\circ} \)

\(\gamma=84^{\circ} \)

\(\delta=180^{\circ} -\alpha=180^{\circ} -72^{\circ} \)

\(\delta=108^{\circ} \)

  !

The sum of the prime factors is 14.

Angle MUV = 25 degrees.

Thanks @Melody I will try doing it myself! 

Not B.  There is no way you are going to get a negative term, when you multiply everything out, if both sets of parentheses contain only plus signs! 

Triangles A and B are congruent.Find the area of triangle A.

Hello Guest!

\(A=\frac{1}{2}\cdot 11\cdot h\)

\(h=\sqrt{16^2-11^2}\)

\(h=11.619\)

\(A=\frac{1}{2}\cdot 11\cdot 11.619\)

\(A=63.9\)

  !

Find the area of the triangle bounded by the y-axis and the lines y - 3x = -2 and 3y + x = 12.

Hello Guest!

\(y=3x-2\\ y=-\frac{1}{3}x+4\)

\(3x-2=\frac{12-x}{3}\\ 9x-6=12-x\\ 10x=18\)

\(x=1.8\)

\(A=\frac{1}{2}\cdot (4-(-2))\cdot 1.8\)

\(A=5.4\)

  !

Ok apart from the final step.  Should end up with r = 51/8

As follows:

Sorry, I don't know trigonometry yet, is there a geometric solution?

you just work you way down with a factor tree

ex.    280

280

=10*28

=2*5*7*4

=2*2*2*5*7

the prime factors are   2, 5 and 7

The radius, r, in inches, of a spherical balloon is related to the volume, V, by r(V) = cubed root(3V/4pi). Air is pumped into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t.

This is what I get.

\(r(V)=\left(\frac{3V}{4\pi}\right)^{1/3}\qquad V(t)=10+20t\\~\\ r(V(t))=\left(\frac{3(10+20t)}{4\pi}\right)^{1/3}\\~\\ r(V(t))=\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}\\~\\ r(V(t))=10\qquad when\\~\\ \left(\frac{3(5+10t)}{2\pi}\right)^{1/3}=10\\~\\ \frac{3(5+10t)}{2\pi}=1000\\ 5+10t=\frac{2000\pi}{3}\\ 10t=\frac{2000\pi-15}{3}\\ t=\frac{2000\pi-15}{30}\\ t=\frac{400\pi-3}{6} \)

LaTex:

r(V)=\left(\frac{3V}{4\pi}\right)^{1/3}\qquad V(t)=10+20t\\~\\
r(V(t))=\left(\frac{3(10+20t)}{4\pi}\right)^{1/3}\\~\\
r(V(t))=\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}\\~\\
r(V(t))=10\qquad when\\~\\
\left(\frac{3(5+10t)}{2\pi}\right)^{1/3}=10\\~\\
\frac{3(5+10t)}{2\pi}=1000\\
5+10t=\frac{2000\pi}{3}\\
10t=\frac{2000\pi-15}{3}\\
t=\frac{2000\pi-15}{30}\\
t=\frac{400\pi-3}{6}

Suppose five crates are weighed two at a time in all possible ways. the weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. how much does each crate weigh?

Definitely not impossible lol.

1) Start by adding all those up.

2) How many times has each crate been included in total?   Divide by that number to the sum of all of the 5 weights.

3) What is the sum of the 2 lightest ones?  What is the sum of the 2 heaviest ones?  Subtract these from the total and you will have the middle weight.

4) Now juggle the numbers until you get the other 4   

            Hint: the lightest 2 will be   55-alpha, and 55+alpha

Have fun :)

Why have you all done this person's homework.

Wouldn't they have more chance of learning something if you just gave hints? 

I suspect this is meant to be as follows

130.25 cm² = 1.3025 dm² 

A = 7 * 9 + 7 * 4 + ( 5² * 3.14 ) / 2 = 130.25 cm²

Draw a right-angled triangle with an acute angle theta.

Let                     tan(theta)=x/2

Now find             sec(theta)

Lastly think about the possible signs   +/-