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kolbaska11 [484]
3 years ago
6

Solve each equations for x. How are these equations similar?

Mathematics
1 answer:
Artyom0805 [142]3 years ago
4 0
If you multiply a. by x it is equivalent to b.
So both are x= z/(w+x)
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Csc2theta=csctheta/2costheta<br><br>Can you help verify the identity
Yakvenalex [24]

csc(2x) = csc(x)/(2cos(x))

1/(sin(2x)) = csc(x)/(2cos(x))

1/(2*sin(x)*cos(x)) = csc(x)/(2cos(x))

(1/sin(x))*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)/(2*cos(x)) = csc(x)/(2cos(x))

The identity is confirmed. Notice how I only altered the left hand side (LHS) keeping the right hand side (RHS) the same each time.

7 0
3 years ago
What is the value of the function at x=−2?
adell [148]

x = -2

function value at x = -2 is intersection of the x value and graph

which is, 3

Hope it is clear:)

5 0
3 years ago
Mr Todd buys 7 drinks at 48p each and 8 drinks at 52p each. What is the total cost of the 15 drinks?
Lesechka [4]

Answer:

80p

if you are adding 4p every time you add a drink you will be adding 7  drinks therefore you have to do (7x4p) this will equal 28p then you have to add 28p to 52p

Hope this helps

6 0
2 years ago
How to work it out logically? Question a !!!
Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

                          =L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha

Area of region 2 = ar(DHBC)

                       =2000\times1500\\\\=3000000m^2=300ha

Area of region 3 = ar(GFEH)

                             (2000+1500)\times 1000\\\\=3500000m^2=350ha

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

                                                 =987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0
3 years ago
Read 2 more answers
Ryan also found that he could model the height of his plant with the equation h = 0.5d + 4, where d is the number of days and h
shusha [124]

Answer:

{4, 4.5, 5, 5.5, 6, 6.5, 7} {0, 1, 2, 3, 4, 5, 6}. 4 ≤ h ≤ 7 0 ≤ d ≤ 6. 4 ≤ d ≤ 7. {4, 5, 6, 7}. 0 ≤ h ≤ 6 0 ≤ h ≤ 7

4 0
2 years ago
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