Need help with it I don’t know how to do it

Two sides of a triangle are 7 and 11 the length of the third side x is expressed

slava [35]

Answer:

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ... No; The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: No; 9.

8 0

2 years ago

Needed help for my math please

vesna_86 [32]

Answer:

see explanation

Step-by-step explanation:

(a)

$\frac{a}{a}$ = 1 ( any value divided by itself = 1 )

(b)

$\frac{a}{1}$ = a ( any value divide by 1 is the value itself )

(c)

$\frac{a}{b}$ × $\frac{c}{d}$ = $\frac{ac}{bd}$

The product of 2 fractions is the product of the numerators divided by the product of the denominators

(d)

$\frac{a}{b}$ ÷ $\frac{c}{d}$ = $\frac{ad}{bc}$

To divide 2 fractions, leave the first fraction, change division to multiplication and turn the second fraction upside down, that is

$\frac{a}{b}$ ÷ $\frac{c}{d}$ = $\frac{a}{b}$ × $\frac{d}{c}$ = $\frac{ad}{bc}$

(e)

$\frac{a}{d}$ + $\frac{b}{d}$ = $\frac{a+b}{d}$

Since the fractions have a like denominator, add the numerators leaving the denominator. This applies to subtraction also

(f)

$\frac{a}{d}$ - $\frac{b}{d}$ = $\frac{a-b}{d}$

See explanation for part (e)

6 0

2 years ago

A 6 foot tall football player casts a 16 inch shadow at the same time stadium we discussed at six for shadow how tall the bleach

katrin [286]

You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.

6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.

So the bleachers are 9/2 x 6 ft = 27 ft.

For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.

So the stage will have dimensions 216 times larger than 1.75" by 3".

That would be 31ft 6ins x 54ft.

Live long and prosper.

5 0

2 years ago

Can someone help me with these questions??

Naddika [18.5K]

What questions I mean there is no questions

4 0

2 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

2 years ago