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3 years ago

Evaluate the function. f(x) = -x2 + 5x + 15 Find f(9)

Mathematics

2 answers:

ss7ja [257]3 years ago

6 0

When evaluating this problem F(9) = 42

Keith_Richards [23]3 years ago

5 0

Answer:

f(9) = 42

Step-by-step explanation:

f(9) = (-9)*2 + 5*9 +15

= -18 + 45 + 15

= 42

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What are two rational numbers between -3/4 and -2/3

Vadim26 [7]

The two rational numbers between $- \frac{3}{4}$ and $- \frac{2}{3}$ is $- \frac{32}{48}$ , $- \frac{36}{48}$

<h3>How to find the rational numbers between -3/4 and -2/3?</h3>

In the form of p/q, which can be any integer and where q is not equal to 0, is expressed as rational numbers. As a result, rational numbers also contain decimals, whole numbers, integers, and fractions of integers (terminating decimals and recurring decimals).

given that -3/4 and -2/3

now take L.C.M between these two rational numbers is 12.

now multiply -3/4 with 3 both numerator and denominator

$- \frac{3}{4} \times \frac{3}{3} = - \frac{ 9}{12}$

again multiply -9/12 with 4 both numerator and denominator

$- \frac{9}{12} \times \frac{4}{4} = - \frac{36}{48}$

now multiply -2/3 with 4 both numerator and denominator

$- \frac{2}{3} \times \frac{4}{4} = - \frac{8}{12}$

again multiply -8/12 with 4 both numerator and denominator

$- \frac{8}{12} \times \frac{4}{4} = - \frac{32}{48}$

Hence the -36/48 and -32/48 are rational numbers between -3/4 and -2/3

Learn more about rational numbers, refer:

brainly.com/question/12088221

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1 year ago

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AVprozaik [17]

Answer:

3cm

Step-by-step explanation:

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2 years ago

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Pepsi [2]

Answer:

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Reason:

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3 years ago

Simplify the expression.

Anna35 [415]

6,372 I used a calculator, I hope this helps

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3 years ago

Read 2 more answers

Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams

hram777 [196]

Answer:

$X(16)=25.71grams$

Step-by-step explanation:

let $X(t)$ denote grams of $C$ formed in $t$ mins.

For $X grams$ of $C$ we have:

$\frac{2}{3}Xg$ of A and $\frac{1}{3}Xg$ of B

Amounts of A,B remaining at any given time is expressed as:

$40-\frac{2}{3}Xg$ of A and $50-\frac{1}{3}Xg$ of B

Rate at which C is formed satisfies:

$\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}$

Apply the initial condition, $X(0)=0$ ,to the expression above

$0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}$

Now at $X(8)=15$ :

$15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778$

Substitute in X(t) to get

$X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71$

5 0

2 years ago