2 years ago

Q6. . Make t the subject of the formula w = 3t + 11

Mathematics

1 answer:

Usimov [2.4K]2 years ago

3 0

Answer:

t = w/3-11/3

Step-by-step explanation:

w = 3t + 11

3t = w-11

t = 1/3(w-11

t = w/3-11/3

You might be interested in

Help me with these asap

IgorLugansk [536]

Use the app that you can easily find answers on

3 0

2 years ago

1/2 of 1/2 of 56 is A. 26 B. 27 C. 28 D. 29A. 26 B. 27 C. 28 D. 29

Vinil7 [7]

The answer would be C.28

5 0

2 years ago

4h+7=4h-3 i think i have it but im not sure 100% so i need hlp

SSSSS [86.1K]

<u>4h + 7 = 4h - 3</u>

Subtract 4h from each side: 7 = -3

There is no value of ' h ' that can make this a true statement,
so the equation has no solution.

5 0

2 years ago

Read 2 more answers

6 hundred thousands + 38 hundreds

Mekhanik [1.2K]

600,000 + 3,800 = 603,800

3 0

3 years ago

Find the polynomial f(x) of degree 3 with real coefficients that has a y-intercept of 60 and zeros 3 and 1+3i.

Sauron [17]

$\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0$

[ correction added, Thanks to @stef68 ]

$\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)$

5 0

3 years ago