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3 years ago

How would I solve this?

Mathematics

2 answers:

WINSTONCH [101]3 years ago

6 0

This is the answer ^^

goldenfox [79]3 years ago

3 0

So you can convert them into top heavy fractions 5/2 and 14/3 then you need to make the denominators(the bottom numbers) the same
So they both go into 6
15/6+28/6=43/6 or 7 1/6

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11.<br>The domain of f(x) is [4, 5, 6]. If f(x) = 3x + 2, what is the range?

grigory [225]

Answer:

The range of the function is:

Range R = {14, 17, 20}

Step-by-step explanation:

Given the function

$f(x) = 3x + 2$

We also know that range is the set of values of the dependent variable for which a function is defined.

In other words,

Range refers to all the possible sets of output values on the y-axis.

We are given that the domain of the function is:

Domain D = {4, 5, 6}

Now,

substituting x = 4 in the function

f(4) = 3(4) + 2

f(4) = 12 + 2

f(4) = 14

substituting x = 5 in the function

f(5) = 3(5) + 2

f(5) = 15 + 2

f(5) = 17

substituting x = 6 in the function

f(6) = 3(6) + 2

f(6) = 18 + 2

f(6) = 20

Thus, we conclude that:

at x = 4, y = 14

at x = 5, y = 17

at x = 6, y = 20

Thus, the range of the function is:

Range R = {14, 17, 20}

3 0

3 years ago

3(5z - 7) - 2 (9z - 11) = 4 (8z - 13) - 17

marysya [2.9K]

Answer z=2

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2 years ago

Josie has 47$ left on her checking account if she writes a check for 55$ what will josies balance be

HACTEHA [7]

She will have negative eight dollars. Thats all I can say

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3 years ago

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.

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3 years ago

If the three angles of a quadrilateral are 120, 130, 100 then what is the fourth angle?

nalin [4]

Answer:

c. 40

Step-by-step explanation:

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3 years ago