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2 years ago

Please answer one of these questions

Mathematics

1 answer:

Tamiku [17]2 years ago

8 0

Answer:

Step-by-step explanation:

black

diamond

picture card

not ace

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What are the real solutions of x squared =225

Gnesinka [82]

The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.

<h3>What are the real solutions of the equation as given in the task content?</h3>

It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;

x² = 225

x = ± 15

Therefore, the real solutions of the equation are; +25 and -25.

Read more on real solutions of equations;

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3 0

1 year ago

If people who bought 4 souvenir spent $2 for each one,how much did all the people who bought 4 souvenir spend in all ?

puteri [66]

I believe it is $0.50.

$2 divided by 4 equals $0.50.

8 0

2 years ago

Solve each triangle. Round your answers to the nearest tenth.<br> m∠A = 60°, m∠B = 30°, c = 30 cm

Svetach [21]

Answer:

mb = 30

Step-by-step explanation:

i got it from me school

8 0

1 year ago

Read 2 more answers

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

3 years ago

Find the equation of a plane that is perpendicular to the vector −4i⃗ −4j⃗ −k⃗ and passing through the point (−2,−5,5)

vredina [299]

hello ....

the equation of a plane that is ; ax+by+cz +d =0

the vector perpendicular to this plane is : V(a,b,c)

in this exercice ; a = -4 b= -4 c = -1

then: the equation of a plane that is ; -4x-4y-z +d =0

but the plane passing through the point (−2,−5,5) :

-4(-2)-4(-5)-(5) +d =0

23+d =0

d =-23

the equation of a plane is : -4x-4y-z-23 =0

5 0

3 years ago