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3 years ago

Please help with these four questions and thanks

Mathematics

1 answer:

alisha [4.7K]3 years ago

4 0

Answer:

Question 19: 4(x+5)+3 = 4·x+4·5+3 (second option).

Question 20: (117.50-83.50)+6(22.50+3) = 34+6(25.50) --> (last option).

Third question:

375=x-28

x= 375+28

x= 403

Fourth question:

0.57, 0.66, 0.75, 0.84,...

0.66-0.57= 0.09 and 0.75-0.66= 0.09 and 0.84-0.75= 0.09

So, the next three terms are: 0.93, 1.02, 1.11

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Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 16cm and a height of 8cm, at the ra

andriy [413]

Let h=height of water
Let r=radius of water surface
r/h=16/8 =2, so r=2h.
The volume of water is:
v=(1/3)×π×r²×h
=(1/3)×π×(2h)²×h
=4/3πh³
dv/dh=4πh^2
By chain rule:
dv/dt=dv/dh×dh/dt
but
dv/dt=4
thus:
4=(4πh)×dh/dt
dh/dt=4/(4πh²)
when h=6cm we have:
dh/dt=4/(4π6²)
=0.00884 cm³/min

8 0

4 years ago

Read 2 more answers

Hello ! :) please help me with a for me to mark you brainlst :)

bixtya [17]

It’s positive!! hope that helps!

6 0

3 years ago

Read 2 more answers

5.) A car dealership has hired you to determine how many sales they can make based on how many flash sales they have in a quarte

Step2247 [10]

Answer:

its B

Step-by-step explanation:

sorry if its wrong

5 0

3 years ago

Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

ikadub [295]

The vertex-form of a quadratic equation is $f(x)=a(x-h)^{2}+k$ , where (h, k) is the vertex of the parabola.

To complete the square of a quadratic expression in standard form, we write the coefficient of x as 2*b, then add and subtract $b^{2}$ .

$f(x)= x^{2} -8x+3$

$f(x)= x^{2} -2*4x+3$

so b=4, now add and subtract $4^{2}$ .

$f(x)= x^{2} -2*4x+ 4^{2}- 4^{2}+3$

$f(x)=( x^{2} -2*4x+ 4^{2})- 16+3$

$f(x)=(x-4)^{2}-13$

3 0

4 years ago

Please Help! will give brainlest!

zzz [600]

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos² $\theta$ in the place of sin² $\theta$ :

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

${x}^{} = \frac{1}{2} \\ x = - 1$

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$

In conclusion,

$\displaystyle\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

7 0

2 years ago