Will someone please help me with this??

Mathematics

1 answer:

ehidna [41]2 years ago

7 0

$\huge \mathfrak { \green{A} \pink{n} \orange{s}{w}{\blue{e}}{ \gray{r}}} \\ \\ \: \: \: \implies \: \mathbb{C. - 3}$

The leading term is the term containing the highest power of the variable: the term with the highest degree. The leading coefficient is the coefficient of the leading term.

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What is f(-3) for the function f(a)=-2a2-5a+4?

posledela

Answer:

-25

Step-by-step explanation:

2a² - 5a + 4, for f(-3)

2(-3)² -5(-3)+4

2*(-27)+15+4

-54+19

= -25

4 0

3 years ago

The volume of gas in a container at a constant temperature varies inversely with the pressure. If the volume is 32 cubic centime

Stells [14]

Answer:

p=32/15 pounds

Step-by-step explanation:

Step 1: Write the correct equation. Inverse variation problems are solved using the equation y=k/x. In this case, you should use v and p instead of x and y. y=k/x---> v=k/p

Step 2: Use the information given in the problem to find the value of k. In this case, you need to find k when v = 32 and p = 4. 32=k/4 128=k

Step 3: Rewrite the equation from step 1 substituting in the value of k found in step 2. v=128/p

Step 4: Use the equation found in step 3 and the remaining information given in the problem to answer the question asked. In this case, you need to find p when v = 60.

60=128/p 60p=128 p=32/15 pounds

8 0

3 years ago

the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t

Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count $Pr(A|B).$

Use formula for conditional probability:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.$

1. The event $A\cap B$ consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $A\cap B$ is

$Pr(A\cap B)=\dfrac{1}{20}.$

2. The event $B$ consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $B$ is