All categories

2 years ago

Find length indicated

Mathematics

1 answer:

VMariaS [17]2 years ago

3 0

Answer:

Step-by-step explanation:

the equation is

33 = 3x-28+3x-30+x

33=7x-28-30

33=7x-58

+58 +58

91=7x

x=13

DE=13

You might be interested in

Lisa has a basket of 17 tomatoes.She makes sauce with 9 tomatoes.If Lisa wants to split up the rest between 3 friends and hersel

givi [52]

17-9=8
8÷3=2.66
Each person gets 2.66 tomatoes

3 0

3 years ago

Read 2 more answers

An artist wants to frame a square painting with an area of 400 square inches . she wants

Artemon [7]

(a) $x^{2}$ = 400
(b) $\sqrt{x^2} = \sqrt{400}$
$x=20,-20$
(c) Only the positive solution makes sense because you cannot have a negative side length.
(d) Since one side of the painting is 20 inches, all four sides together will be 20*4 = 80 inches

6 0

2 years ago

Plz help me this is really hard thx and im giving brainly

shutvik [7]

Answer:

$\pmb{4000+800+10+2 }$

Step-by-step explanation:

4812

4×(1000)+8(100)+10(1)+1(2)

4000+800+10+2

6 0

2 years ago

Read 2 more answers

A circle’s circumference is approximately 76 cm. Find the radius, diameter, and area of the circle to the nearest centimeter. Us

PolarNik [594]

Answer: Area= 459.73cm^2

Step-by-step explanation: Circumference of a circle is 2πr

76=2×3.14×r

r= 76÷(6×3.14)

r=12.10

Area of a circle= πr^2

Area= 3.14×(12.10)^2

Area= 456.73cm^2

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago