PLZ HELP FOR BRAINLIEST !!

A car travels for 4(t+3) hours at a constant speed of 10(t+3) km/h. If the total distance travelled by the car is 810 km, find t

IceJOKER [234]

Step-by-step explanation:

$\because \: time \times speed = distance \\ \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \: \{2(t + 3) \}^{2} = {9}^{2} \\ \therefore \: \{2(t + 3) \} = {9} \\ \therefore \: t + 3 = \frac{9}{2} \\ \therefore \: t = \frac{9}{2} - 3 \\ \therefore \: t = \frac{9 - 6}{2} \\ \therefore \: t = \frac{3}{2} \\ \therefore \: t = 1.5 \: hrs \\$

$\because \: time \times speed = distance \\ \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \: \{2(t + 3) \}^{2} = {9}^{2} \\ \therefore \: \{2(t + 3) \} = {9} \\ \therefore \: t + 3 = \frac{9}{2} \\ \therefore \: t = \frac{9}{2} - 3 \\ \therefore \: t = \frac{9 - 6}{2} \\ \therefore \: t = \frac{3}{2} \\ \therefore \: t = 1.5 \: hrs \\ speed \: of \: car = 10(t + 3) \\ \hspace{60 pt} = 10(1.5 + 3) \\ \hspace{60 pt}= 10 \times 4.5 \\ \red{ \boxed{ \bold{ \therefore \: speed \: of \: car = 45 \:km/ h}}} \\$

3 0

3 years ago

Which equation provides the best estimate for 35.78 ÷ 12.31?

kobusy [5.1K]

Estimate
35.78 = 36
12.31 = 12
so
36/12 = 3

answer
3

5 0

3 years ago

Simplify the following algebraic expression: 2 (2x + 3y) + 3 (5 – 2x) +1+y

Gnoma [55]

Answer: -2x+7y+16

Step-by-step explanation:

1. Distribute 2 through the parentheses

4x+6y+3(5-2x)+1+y
4x+6y+15-6x+1+y

2. Collect like terms

-2x+6y+15+1+y
-2x+7y+15+1
-2x+7y+16

3. Solution, -2x+7y+16

5 0

2 years ago

To which subsets of the real numbers does the number 82 belong?

larisa86 [58]

We know that
82 is an integer, a whole number, a natural number, and a rational number (since 82 = 82/1)

So
82 belongs to the set of integers, set of whole numbers, set of natural numbers, and the set of rational numbers

3 0

3 years ago

Help algebra 2 please

kati45 [8]

Answer:

Step-by-step explanation:

(f*g)(x) = (-5x² + 2x + 7) (x +1)

= x* (-5x² + 2x + 7) + 1*(-5x² + 2x + 7)

= x*(-5x²) + x*2x + x*7 - 5x² + 2x + 7

= -5x³ + 2x² + 7x - 5x² + 2x + 7

= - 5x³ + 2x² -5x² + 7x + 2x +7 {Combine like terms}

= -5x³ - 3x² + 9x + 7

4) (f*g)(x) = (x² + 2x + 4)(x - 2)

= x*(x² + 2x + 4) - 2*(x² + 2x + 4)

= x*x² + x*2x + x*4 - 2*x² - 2*2x -2* 4

= x³ + 2x² + 4x -2x² -4x - 8

= x³ - 8

$2) Speed=\dfrac{distance}{time}\\\\\\= \dfrac{d(h)}{t(h)}\\\\= \dfrac{2\sqrt{h}}{3\sqrt{4h}}=\dfrac{2*\sqrt{h}}{3*2\sqrt{h}}\\\\\\=\dfrac{1}{3}$

$1)d(h) = \sqrt{16h^{4}}=\sqrt{2*2*2*2*h*h*h*h}=2*2*h*h=4h^{2}\\\\\\t(h) = 3h^{4}-3h^{3}-5h^{2}= h^{2}(3h^{2}-3h - 5))\\\\Speed=\dfrac{4h^{2}}{h^{2}(3h^{2}-3h-5)}\\\\=\dfrac{4}{3h^{2}-3h-5}$

7 0

3 years ago