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3 years ago

Please help me get the question right away

Mathematics

2 answers:

dmitriy555 [2]3 years ago

7 0

True I hope you get it right

Feliz [49]3 years ago

5 0

Answer:

true

Step-by-step explanation:

hopefully this helps :)

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The table below shows the distance a car travels and the amount of gasoline left in the tank of the car. Distance Traveled and G

hichkok12 [17]

Answer:

b: 4

Step-by-step explanation:

i took the test on edge 2020

4 0

3 years ago

Read 2 more answers

Ivana works at a department store selling perfume and makeup. She earns $375 per week and 4% commission on her sales. If Ivana's

Aleks04 [339]

Answer:

Step-by-step explanation:

Step one:

given data

weekly earnings= $375

Commission is 4% of sales.= 0.4

supposing sale is x, and the total earning is y

the expression for the total earning will be

y=375+0.4x

Step two:

for x= 2513, y will be

y=375+0.4(2513)

y=375+1005.2

y=1380.2

<em><u>Hence her total earnings for the week will be $1380.2</u></em>

3 0

3 years ago

The image of the polygon ABCDE reflected across the x-axis. A(-1,-1),B(0,1),C(4,2),D(6,0),E(3,-3) The vertices of the image are.

4vir4ik [10]

A reflection across the x-axis has the rule:

(x,y)→(x,-y).

Then:

A(-1,-1)→A'(-1,1),
B(0,1)→B'(0,-1),
C(4,2)→C'(4,-2),
D(6,0)→D'(6,0),
E(3,-3)→E'(3,3).

Answer: the vertices of the image are A'(-1,1), B'(0,-1), C'(4,-2), D'(6,0) and E'(3,3).

6 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>

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