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4 years ago

Express 236.7 miles in 4.5 days as a unit rate

Mathematics

1 answer:

Andre45 [30]4 years ago

6 0

The answer would be 52.6 miles per 1 day. Take 236.7 and divide it by 4.5.

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Kol and Kara solve this problem in two different ways the perimeter of a rectangle is 46 centimeters it’s length is 8 centimeter

JulsSmile [24]

Answer:

15cm

Step-by-step explanation:

because length is given as 8 cm and perimeter is length+width+length+width

so if you take 8cmX2 since its taken twice to get the perimeter and 15cmX2 you get 46cm which originally the question said was the peemiter

5 0

2 years ago

Read 2 more answers

Select the two values of x that are roots of this equation x^2-5x+2=0

Nataly [62]

The roots of the equation is x = 4.56 OR x = 0.44

<h3>Quadratic equation</h3>

From the question, we are to determine the roots of the given equation

The given equation is

x² -5x +2 = 0

Using the formula method,

$x =\frac{-b \pm \sqrt{b^{2}-4ac } }{2a}$

In the given equation,

a = 1, b = -5, c = 2

Putting the values into the formula,

$x =\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(2) } }{2(1)}$

$x =\frac{5 \pm \sqrt{25-8} }{2}$

$x =\frac{5 \pm \sqrt{17} }{2}$

$x =\frac{5 + \sqrt{17} }{2}$ OR $x =\frac{5 - \sqrt{17} }{2}$

$x =\frac{5 + 4.12}{2}$ OR $x =\frac{5 - 4.12}{2}$

$x =\frac{9.12}{2}$ OR $x =\frac{0.88}{2}$

x = 4.56 OR x = 0.44

Hence, the roots of the equation is x = 4.56 OR x = 0.44

Learn more on Quadratic equation here: brainly.com/question/8649555

#SPJ1

3 0

2 years ago

Help me with this please

snow_tiger [21]

B) y=34.84x-2054 I believe that is the right answer

7 0

3 years ago

39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t

Katyanochek1 [597]

Write tan in terms of sin and cos.

$\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}$

Recall that

$\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1$

Rewrite and expand the given limand as the product

$\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)$

Then using the known limit above, it follows that

$\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}$

4 0

2 years ago

The shadows of two vertical poles were measured at the same time to be 6m and 15m long. If the first pole is 8cm long, find the

son4ous [18]

Answer:

20 cm

Step-by-step explanation:

Shadow of vertical poles : 6m ; 15m

Length of poles : 8cm ; h

8cm long pole = shadow length, 6m

h cm long pole = shadow length, 15 m

Using cross multiplication :

8cm * 15 m = h cm * 6m

120 = 6h

h = 120 / 6

h = 20 cm

Height of second pole = 20cm

8 0

3 years ago