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2 years ago

Please Help! Knowing the negative sign (–) also means "opposite", how do you we read this expression, -8 • 2?

Mathematics

2 answers:

ycow [4]2 years ago

7 0

The answer is -12468

Nostrana [21]2 years ago

3 0

Negative eight times two? not quite sure what you are asking

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Solve for x, y, and all of the missing angles. d = (2y+2) c = (3y + 18) a 50° b = (2x + 4) b degrees b degrees . degrees d = deg

CaHeK987 [17]

Answer:

that answer was wrong your dum

Step-by-step explanation:

4 0

2 years ago

The shirts cost a $150 for a 100 shirts how much per shirt

Irina18 [472]

Please mark brainliest

Step-by-step explanation:

1.5 dollars a shirt

3 0

2 years ago

A store is having a sale on trail mix and jelly beans. For 6 pounds of trail mix and 5 pounds of jelly beans, the total cost is

-Dominant- [34]

6t + 5j= 34
2t + 3j = 16

12t + 10 j = 68
12t + 18j = 96
-8j = -28
j = 3.5

jelly beans= 3.5 and trail mix = 2.75

8 0

3 years ago

Help! Best answer gets brainliest

julsineya [31]

Answer:

7.9

Step-by-step explanation:

the first thing to do is find the second projection.

16-7 = 9

than we can use a proportion to find BD

7 : BD = BD : 9

BD = √9 x 7 = √63 = 7.9

we have to use the square root because in the proportion BD Is repeated two times. If we don’t use it, we find BD^2

7 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

2 years ago