7. One half of a cantaloupe was shared

PLS HELP IT’S DUE TMR!!

ValentinkaMS [17]

Answer:

Use ≈ (approximately equal) sign as the scientific notation.

Step-by-step explanation:

(a) 0.001872 ≈ 0.0019

(b) 0.3411 ≈ 0.34

(c) 0.000845 ≈ 0.00085

*Zeros before non-zero numbers are not significant.

*Zeros appearing between two non-zero digits are significant.

Hope this helps!!

6 0

2 years ago

Which points on the graph represent values that could be in the table? A B C D

Dafna11 [192]

Answer:

uhhh

Step-by-step explanation:

uh I don't see the graph

4 0

4 years ago

Read 2 more answers

Determine the equation of the polynomial, p(x),in both factored and standard form.

melamori03 [73]

We can see that the graph touches $x=-1$ without crossing the x-axis (i.e. it is a double solution), and then there's another zero at $x=2$ (this time it's a crossing zero, so a single solution).

This leads, up to multiple, to the polynomial

$p(x)=a(x+1)^2(x-2)$

If we impose the passing through $(0,4)$ we have

$p(0)=4=a(1)(-2) \iff -2a=4 \iff a=-2$

So, the polynomial is

$p(x)=-2(x+1)^2(x-2)=-2 x^3 + 6 x + 4$

Finally, to solve $p(x)$ , simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if $x>2$ , so that's the solution to your question.

4 0

3 years ago

During 4 days, the price of the stock of PEV Corporation went up 1/4 of a point, down 1/3 of a point, down 3/4 of a point, and u

HACTEHA [7]

The price of the stock of PEV Corporation varies as shown. we have to determine the net charge. Net change is just how much it changed overall, so we will look for all the ups minus all the downs.

Here, "went up" goes with a + sign, and "went down" goes with a - sign.

So, Net charge = $\frac{1}{4} - \frac{1}{3}-\frac{3}{4} + \frac{7}{10}$

= $\frac{1}{4}-\frac{3}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{2}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{1}{2}- \frac{1}{3} + \frac{7}{10}$

LCM of 2,3 and 10 is 30

= $\frac{-15-10+21}{30}$

= $\frac{-4}{30}$

= $\frac{-2}{15}$

So, the net charge is $\frac{-2}{15}$

8 0

3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia

ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

Xa = Xb

-6 + (3*t) = 7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

(6.8*t) = 13.4

t = 13.4/6.8

t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0

3 years ago