All categories

3 years ago

Given F ( x ) = -2/3 x - 4 What is the zero of this function?

Mathematics

1 answer:

sineoko [7]3 years ago

5 0

Answer:

-4

Step-by-step explanation:

f(0)=(-2/3)(0) - 4

= - 4

You might be interested in

THERES A PIC!! pls answer!! will give brainliest!!

Andrew [12]

Answer:

The slope/rate of change/growth is 5

Step-by-step explanation:

(-8--3)/(-2--1)=-5/-1=5

7 0

3 years ago

Using the numbers 1 through 9 only once in the boxes, what is the largest value of x you can get?

aliina [53]

Using the numbers 1 through 9 only once in the boxes, the largest value of x you can get is <u>7</u>.

<h3>What are mathematical operations?</h3>

Mathematical operations refer to the calculation of values using operands and mathematical operators.

Mathematical operations are formed like functions to take zero or more input values (operands) to a well-defined output value.

Some of the mathematical operators perform addition, subtraction, multiplication, division, exponentiation, and modulus operations.

<h3>Data and Calculations:</h3>

Mx + y = z

Let m = 1

x = 7

y = 2

1 x 7 + 2 = 9

7 + 2 = 9

9 = 9

Thus, using the numbers 1 through 9 only once in the boxes, the largest value of x you can get is <u>7</u>.

Learn more about mathematical operations at brainly.com/question/20628271

#SPJ1

6 0

1 year ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago

On Monday a team of street sweepers cleaned 1/3 of the city block. Tuesday, the team cleaned 1/2 as many blocks as on Monday. Ho

IrinaVladis [17]

Answer:

1/2 half

Step-by-step explanation:

(1/3)/2 = 1/6

1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

6 0

3 years ago

Vaselesa [24]

Answer:

75(1/3)^(n-1).

Step-by-step explanation:

This is a Geometric sequence whose common ratio is 15/75 3/15 = 1/3.

so the nth term an = a1 r^(n-1)

= 75(1/3)^(n-1).

6 0

3 years ago