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4 years ago

Write the function in function notation: y=7b+8

Mathematics

1 answer:

kow [346]4 years ago

3 0

You would just write f(x) (f of x) in for y: f(x) = 7x + 8.
Hope it helps.

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Rewrite the equation 2 x + 3 y = 6 in slope-intercept form I will give brainliest :-)

Vladimir [108]

Step-by-step explanation:

2x + 3y = 6

3y = -2x + 6

divide through by 3

y = (-2/3)x + 6............1

but equation of a straight line is

y = mx + c..............2

compare equation 1&2, m = -2/3 and c = 6

where m = gradient

c = intercept

6 0

3 years ago

the common ratio in geometric series is 0.5 and the first term is 256. what is the sum of the first 6 terms in the series?

melisa1 [442]

Answer:

501.76

Step-by-step explanation:

common ratio r = 0.5

first term a = 256

sum of G.p = Sn= a( r^n -1)/r-1

sum of the first 6 terms =

S6 = 256( 0.5^6 - 1)/ 0.5 - 1

= 256( 0.0156 - 1)/-0.5

= 256( -0.98)/-0.5

= - 250.88/-0.5

S6 = 501.76

7 0

3 years ago

Elena starts her new job working at an office supply store. She works for 6 hours and makes a total of $54.

docker41 [41]

Answer:

I believe the answer would be A.

Step-by-step explanation:

A says 6 x h is 54.

^ ^ ^

important!

6 is the number of hours.

h is how much money made per hour.

54 is the total after working 6 hours.

4 0

3 years ago

Help me with math pls 20 pints show work !

eduard

Answer:

the answer is 16x

Step-by-step explanation:

first, you have to add like terms so 6x+10x then you get your answer

5 0

3 years ago

A man is in a boat 2 miles from the nearest point on the coast. He is to go to point Q, located 3 miles down the coast and 1 mil

mafiozo [28]

Answer:

The answer is "0.45385".

Step-by-step explanation:

The time is taken to reach the coast $= \frac{\sqrt{(2^2 + x^2)}}{4}$

The time is taken to reach Q after reaching the coast $= \frac{\sqrt{(1 + (3-x)^2)}}{4}$

Total time, $T= \frac{\sqrt{(1 + (3-x)^2)}}{4}+ \frac{\sqrt{(1 + (3-x)^2)}}{4}$

This has to be minimum,

$\to \frac{dt}{dx} = 0$

$\to \frac{\sqrt{(2^2 + x^2)}}{2} + \frac{\sqrt{1 + (3-x)^2}}{3}\\\\ \frac{x}{4}\sqrt{(4+x^2)} + \frac{(x-3)}{(4\sqrt{(x^2-6x+10)}} = 0\\\\\frac{x^2}{16} \times (4+x^2) = \frac{(x-3)^2}{16 \times (x^2-6x+10)}\\\\x^2(4+x^2) = \frac{(x-3)^2}{(x^2-6x+10)}\\\\4x^2+x^4 = \frac{ x^2+9-6x}{(x^2-6x+10)}\\\\ 4x^2+x^4(x^2-6x+10) = x^2+9-6x\\\\4x^4-24x^3+40x^2+x^6-6x^5+10x^4= x^2+9-6x\\\\x^6-6x^5+ 14x^4-24x^3+39x^2+6x-9=0\\\\$

x=0.45385

6 0

3 years ago