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3 years ago

6

5 is 1/10 as much as 500

1 answer:

Sergio [31]3 years ago

5 0

No 5 is 1/100 as much as 500, not 1/10 as much as 500

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Mrs. James displayed the factor tree at the right complete the factor tree to find the number that has prime factorization of 2

Taya2010 [7]

Download the guath math app it will help you with all your math work

4 0

3 years ago

Read 2 more answers

What is the domain of the function.....<br> {TIMED PLS HURRY}

timofeeve [1]

try c im sorry if im wrong i tried

6 0

3 years ago

jan, mya, and Sara ran a total of 64 miles last week. jan and mya ran the same number of miles. Sara ran 8 less miles than mya.

Debora [2.8K]

So,

We'll just use A to represent both Jan and Mya's miles, since they ran the same number.

We have the equations:
1. Jan (J) = Mya (M)
2. Sara (S) = M - 8
3. 2A + S = 64
J = M
S = M - 8

We'll just use A to represent both J and M.

S = M - 8

We'll use Elimination by Substitution.
2A + A - 8 = 64

Collect Like Terms
3A - 8 = 64

Add 8 to both sides
3A = 72

Divide both sides by 3
A = 24

Since
A = J
and
A = M
and
J = M
then
J = 24
M = 24

Substitute
S = 24 - 8
S = 16

Check
24 + 24 + 16 = 64
64 = 64 This checks.

So,
J = 24
M = 24
S = 16

4 0

3 years ago

Read 2 more answers

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago

F(9) if f (x)=5x-16, what is the answer to the math problem

Anastasy [175]

F(9)
sub 9 forx
f(9)=5(9)-16
f(9)=45-16
f(9)=29

f(9)=29

3 0

3 years ago