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3 years ago

If a shape has 4 right angles, then it is a rectangle.

Mathematics

2 answers:

shusha [124]3 years ago

7 0

Answer: leahcool08 is right

Step-by-step explanation:a rectangle has two even size the other two even sides a square has four even sides and square is a rectangle but a rectangle isn't a square

almond37 [142]3 years ago

3 0

a rectangle has two even size the other two even sides a square has four even sides and square is a rectangle but a rectangle isn't a square

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HELP ASAP WILL GIVE BRAINLIEST AND 30 POINTS!!!

Stolb23 [73]

Answer is C qwertyuiopoiuytrewertyuioiuytrewerty

8 0

3 years ago

Solve the following logarithmic equations.<br> log[(x^2 + 2x − 3)^4] = 0

saw5 [17]

Answer:

The solutions are x = 1.24 and x = -3.24

Step-by-step explanation:

Hi there!

First, let´s write the equation:

log[(x² + 2x -3)⁴] = 0

Apply the logarithm property: log(xᵃ) = a log(x)

4 log[(x² + 2x -3)⁴] = 0

Divide by 4 both sides

log(x² + 2x -3) = 0

if log(x² + 2x -3) = 0, then x² + 2x -3 = 1 because only log 1 = 0

x² + 2x -3 = 1

Subtract 1 at both sides of the equation

x² + 2x -4 = 0

Using the quadratic formula let´s solve this quadratic equation:

a = 1

b = 2

c = -4

x = [-b± √(b² - 4ac)]/2a

x = [-2 + √(4 - 4(-4)·1)]/2 = 1.24

and

x = [-2 - √(4 - 4(-4)·1)]/2 = -3.24

The solutions are x = 1.24 and x = -3.24

Have a nice day!

4 0

3 years ago

the volume of the rectangular prism shown below is 96 cubic feet. The dimensions of the prism base are 4 feet and 3 feet. What i

g100num [7]

Height=volume:base area
96:(4*3)= 96:12=8

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

3 years ago

Plz answer with explanation will give brainliest

maksim [4K]

Answer:

B: |8| + (-7)

Step-by-step explanation:

When you are looking for the distance between two numbers, always add. The distance between 8 and -7 is 15 (|8| = 8). If, for instance, the answer was negative, which it is now, just find the absolute value of -15 which equals 15. The answer is B.

3 0

3 years ago