All categories

3 years ago

Please help and find the missing side x

Mathematics

2 answers:

faltersainse [42]3 years ago

7 0

$\text{Hello there!}\\\\\text{A simple way to find the value of x is to find the scale factor}\\\\\text{When you flip the shapes to make it look exactly the same,}\\\text{you would see that the numbers are divisible to it}\\\\\text{Side 21 is congruent to side 3}\\\\\text{Divide 21 by 3 and you'll get 7 (1/7 sine we're going to a smaller shape)}\\\\\text{The x side is congruent to the 91 side}\\\\\text{Divide 91 by 7}\\\\91\div7=13\\\\\large\boxed{x=13}$

mars1129 [50]3 years ago

6 0

The missing side x would be 13. I figured this out by seeing how to get 21 from 3 which is 3 •7 then I did the same with all the others sides and they all added up to __•7 so all you have to do is divide 91 by 7 and you will get 13.

I hope this helped you understand!!:):) Sorry if you couldn’t understand how I explained it

You might be interested in

A pump can fill a swimming pool in 8 hours. The pool also has a drain that can empty the pool in 10 hours. If someone turns on t

cluponka [151]

Answer:

40 hours will it take for the pool to fill.

Step-by-step explanation:

A pump can fill a swimming pool in 8 hours.

Work done by pump to fill in 1 hour is $\frac{1}{8}$

The pool also has a drain that can empty the pool in 10 hours.

Work done by pump to drain in 1 hour is $\frac{1}{10}$

If someone turns on the pump to fill the pool, but forgets to shut the drain.

Work done by both pipe in 1 hour is

$W=\frac{1}{8}-\frac{1}{10}$

$W=\frac{10-8}{80}$

$W=\frac{2}{80}$

$W=\frac{1}{40}$

Both pipe filled $\frac{1}{40}$ part of pool in hours = 1

Both pipe filled complete pool in hours = $\frac{1}{\frac{1}{40}}=40$

Therefore, 40 hours will it take for the pool to fill.

7 0

3 years ago

3a-6b/3ab<br><br> Steps please. Not sure if I originally did this question correctly.

crimeas [40]

Answer:

a-2b/ab

Step-by-step explanation:

3a-6b/3ab

Factorize

3(3a-6b/3ab)

= a-2b/ab

6 0

3 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$

$= - \dfrac{1}{2} cos (4) + \dfrac{1}{2} (1) \\ \\ = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\ = \mathbf{0.82682}$

5 0

3 years ago

Please solve for n and show your work.<br> <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="

k0ka [10]

$\frac{1}{3} (n + 15) = -2$

First, replace the numerator with the parentheses.
$\frac{n+15}{3} =-2$
Second, multiply both sides by 3.
$n + 15 = -2 * 3$
Third, multiply 2 × 3 to get 6.
$n + 15 = -6$
Fourth, take away 15 from each side.
$n = -6 - 15$
Fifth, simplify -6 - 15 to get -21.
$n = -21$

Answer: n = -21

3 0

3 years ago

I sthe answer A? IM only half sure??

trasher [3.6K]

You are correct there is no pattern

5 0

3 years ago

Read 2 more answers