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thank you

To be a binomial, the answer can have only two terms.

It will have a t² term (because 7t·5t = 35t²) and it will have a constant term because, according to the

directions, you can't use zero in the box.

Therefore, the t-term must disappear.

Using an 'x' for the term in the box, we have:  (7t - 10)(5t + x)  =  35t² - 50t + 7xt - 50x.

We need to get  -50t + 7xt  to be zero:  -50t + 7xt  =  0

(dividing out the t)                                    -50 + 7x  =  0

                                                                          7x  =  50

                                                                            x  =  7 1/7

I don't know, Take a guess!

The lower left-hand corner angle of the triangle is 85^o because it is vertical to an 85^o angle.

Since the lines are parallel, the angle supplementary to the lower right-hand corner angle of the 

triangle is 98^o.

This makes the lower right-hand corner angle of the triangle 180^o - 98^o  =  82^o.

Now, we know two of the three angles of the triangle; all we need to do is find the size of the

third angle.  180^o - 85^o - 82^o  =  .......

There seems to be missing information -- are "6" and "-3" consecutive terms, or are there terms

between them that are missing? If so, how many?

Omg!!!!!!!!!!!GUYS I GOT THE ANSWER TO THE WORLD'S TOUGHEST QUESTION. Guest a=9. THANK YOU EVERYONE FOR READING PLS LIKE.

Well Guest, if 3^2 = 9, and 3^2 = a, that means 9 = a.

Therefor we have \(9=a\)

I am assuming 72

Using Chinese Remainder Theorem with Modular Multiplicative Inverse which are incorporated into this short computer code, we get:

i=0;j=0;m=0;t=0;a=(3, 5,7, 9);r= (2, 4, 6, 8);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

315 m +  314, where m =0, 1, 2, 3.......etc.

Then the smallest positive integer = 314

See the solution by heureka here:  https://web2.0calc.com/questions/i-need-help-on-these-problems-asap

Why would that be the case?

Diameter  FG = 68 units 

The angle FEB = 68 degrees because the angle BFE is the right angle 

I wrote "12 (By shoelace formula" thus i wanted checking using this method. sry if it was not that clear!

Something's wrong with this question 

According to the information provided, the lengths of the sides of a triangle XYZ are:   35,  51, and 32

Neither of these 3 angles has a measure of 21º 

Sides: a = 35   b = 51   c = 32

Area: T = 553.042
Perimeter: p = 118
Semiperimeter: s = 59

Angle ∠ A = α = 42.668° = 42°40'5″ = 0.745 rad
Angle ∠ B = β = 99.041° = 99°2'28″ = 1.729 rad
Angle ∠ C = γ = 38.291° = 38°17'28″ = 0.668 rad

Height: ha = 31.602
Height: hb = 21.688
Height: hc = 34.565

I do not think that triangle is possible.

I calculate that Angle YZX must equal  42.66 degrees.   (rounded)

Hi Melody,

Yes, I know I can find its area just using the number plane.

I was trying to apply "Shoelace formula" with simple polygons like this triangle to see if it works or not (Of course there is easier ways for this one!). But when it comes to polygons with 5 vertices...

So if someone can let me know how to apply it correctly ,thx!

No it is not.

Draw the triangle on a rough number plane.

Then use the formula

area of a triangle =  half the base times the perpendicular height to find the answer.

Get back to us with your answer and maybe your working too.

To other answerers:

Please do not give more information.

I dont understand how to do this. Please show steps

EC=

\(\text{Let $ES=ER=30$} \\ \text{Let $DT=DR=17$} \\ \text{Let $TC=SC=32$} \\ \text{Let the perimeter is $c+d+e$} \)

\(\begin{array}{|rcll|} \hline c+d+e &=& 2*17+2*30+2*32 \\ \mathbf{c+d+e} &=& \mathbf{158} \\ \hline \end{array}\)

The perimeter of \(\triangle DEC\) is \(\mathbf{158}\)

calculate angle NAS