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4 years ago

True or false: In (r, theta), the value of r can be negative.

1 answer:

3 0

The answer to the mathematics question presented above is 'True'. It is correct to say that in (r theta), the value of r can be negative. A negative radius can be used when it comes to graphing a "polar'' function. Thus, the answer to the question is 'true'.

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Rani scores 4 times as many baskets as Layla. Tash scores 8 baskets less than Rani. They score 100 baskets in total. How many ba

Sever21 [200]

Answer:

Tash scores 40 baskets

Step-by-step explanation:

Rani = R

Layla = L

Tash = T

R = 4L

T = 4L-8

As L + R + T = 100

L + 4L + 4L-8 = 100

Collect like terms

9L - 8 = 100

+ 8 both sides

9L = 108

divide by 9 both sides

L = 108 / 9

L = 12

Now..... T = 4L-8 substitute L with 12

T = 48 - 8

T = 40

8 0

3 years ago

3 1/7 x 10 answer plz

Yanka [14]

Answer:

31.43

Step-by-step explanation:

3 1/7 = 3.14285714285

3 1/7 x 10 = 31.4285714286 which is about 31.43

4 0

3 years ago

Read 2 more answers

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):