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3 years ago

11

Elena a cumpărat de la cofetărie 12 prajituri unele cu ciocolata si altele cu frișcă numărul prăjiturilor cu ciocolata este cu 2

mai mare decât al celor cu frișcă Câte prăjituri cu frișcă a cumpărat elena

1 answer:

butalik [34]3 years ago

6 0

A cumparat 5 prajituri cu vanilie si 7 cu ciocolata.

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You have a large container of olive oil. You have used 25 quarts of oil.

Svetradugi [14.3K]

Step-by-step explanation:

25 plus 25 equal 50 so 50 percent

7 0

3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$

7 0

3 years ago

Which are vertical angles?<br> O AFE and BFD<br> O BFC and DFE<br> O AFE and CFD<br> O BFC and EFA

mel-nik [20]

the correct answer is a, because a vertical angle is made of 2 intercepting lines.

3 0

3 years ago

Trouble finding arclength calc 2

kiruha [24]

Answer:

$S\approx1.1953$

Step-by-step explanation:

So we have the function:

$y=3-x^2$

And we want to find the arc-length from:

$0\leq x\leq \sqrt3/2$

By differentiating and substituting into the arc-length formula, we will acquire:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx$

To evaluate, we can use trigonometric substitution. First, notice that:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx$

Let's let y=2x. So:

$y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx$

We also need to rewrite our bounds. So:

$y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0$

So, substitute. Our integral is now:

$\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Let's multiply both sides by 2. So, our length S is:

$\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

$y=a\tan(\theta)$

Substitute 1 for a. So:

$y=\tan(\theta)$

Differentiate:

$y=\sec^2(\theta)\, d\theta$

Of course, we also need to change our bounds. So:

$\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0$

Substitute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta$

The expression within the square root is equivalent to (Pythagorean Identity):

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta$

Simplify:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta$

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

$u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta$

And:

$dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)$

Integration by parts:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)$

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)$

Distribute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)$

Now, let's make the single integral into two integrals. So:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Distribute the negative:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

$\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Divide the second and third equation by 2. So: $\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))$

Therefore, our arc length will be equivalent to:

$\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Divide both sides by 2:

$\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Evaluate:

$S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))$

Evaluate:

$S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))$

Simplify:

$S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}$

Use a calculator:

$S\approx1.1953$

And we're done!

7 0

3 years ago

A washer and a dryer cost $917 combined. The washer costs $67 more than the dryer. What is the cost of the dryer?

mart [117]

917 - 67 = 850 (subtract the difference)
850 / 2 = 425 (divide the cost of the washer/dryer)
425 + 67 = 492 (add the difference)
The dryer costs $492.

4 0

3 years ago