Explain why the net shown here cannot be the net of a cube

A student claims that 8^3 x 8^-5 is greater than 1. Explain whether the student is correct or not.

nlexa [21]

Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m

8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1

Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.

7 0

3 years ago

Hi I need help please

Nadusha1986 [10]

Answer: 14 7/8

Step-by-step

I put the 4 1/4 into an improper fraction then did keep change flip

4 0

3 years ago

Read 2 more answers

Write 2740 in scientific notation.

juin [17]

Answer:

2.74 x 10 square root of 3

Step-by-step explanation:

hope this helps give brainliest

4 0

3 years ago

Read 2 more answers

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

3 years ago

I need help with premetier and area for 7 8 9

SCORPION-xisa [38]

The answers to the three shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

<h3>What is the perimeter of a shape?</h3>

Perimeter is the outside boundary of a plane shape.

Analysis:

for figure 7, perimeter = s + s + s = 3s = 3(5.4) = 16.2mm

Height of triangle = $\sqrt{((5.4)^{2} - (2.7)^{2} }$ = 4.68

Area of triangle = 1/2 base x height = 1/2 x 5.4 x 4.68 = 12.64 $mm^{2}$

For figure 8, perimeter = b + b + a = 5.9 + 5.9 + 3.4 = 15.2 inches

Height of triangle = $\sqrt{(5.9)^{2} - (1.7)^{2} }$ = 5.65inches

Area of triangle = 1/2 base x height = 1/2 x 3.4 x 5.65 = 9.61 $inches^{2}$

For figure 9, perimeter = b + a + c = 8.2 + 4.1 + 9.17 = 21.47yards

Area of triangle = 1/2 x base x height = 1/2 x 8.2 x 4.1 = 16.81 $yards^{2}$

In conclusion, the perimeter and area of the given shapes are:

Figure 7; perimeter = 16.2mm, area = 12.64 square mm

Figure 8; perimeter = 15.2inches, area = 9.61 square inches

Figure 9; perimeter = 21.47yards, area = 16.81 square yards

Learn more about perimeter of plane shapes: brainly.com/question/2569205

#SPJ1

5 0

2 years ago