Mr. Park is using the following

Solve each system of equations; nonlinear algebra 2<br><br> x2 + x + y - 26 = 0<br> x + y = 1

BigorU [14]

Answer:

(5,-4) and (-5,6)

Step-by-step explanation:

Given:

$\left\{\begin{array}{l}x^2+x+y-26=0\\ \\x+y=1\end{array}\right.$

Solve it. First, express y in terms of x from the second equation:

$y=1-x$

Substitute it into the first equation:

$x^2+x+1-x-26=0\\ \\x^2-25=0\\ \\(x-5)(x+5)=0$

Apply zero product property:

$x-5=0\ \text{or}\ x+5=0$

So,

$x=5\ \text{or}\ x=-5$

When $x=5,$ then $y=1-5=-4$

When $x=-5,$ then $y=1-(-5)=6$

We get two solutions: (5,-4) and (-5,6)

8 0

4 years ago

Three semicircles are placed in a rectangle, as shown below. The width of the 3 rectangle is 15 . Find the area of the shaded re

Leni [432]

Answer:

Step-by-step explanation:

Width of rectangle = 15 cm

width of rectangle is equal to the radius of each semicircle.

radius of each semicircle (r) = 15 cm.

Diameter of each semicircle = 30 cm

Length of rectangle = 3 x 30 = 90 cm.

$Area \: of \: shaded \: region \\ = Area \: of \: rectangle \\ \: \: \: \: \: - 3 \times area \: of \: one \: semicircle \\ = l \times w - 3 \times \frac{\pi {r}^{2} }{2} \\ \\ = 90 \times 15 - 3 \times \frac{3.14 \times {15}^{2} }{2} \\ \\ = 1350 - 3 \times \frac{3.14 \times 225 }{2} \\ \\ = 1350 - \frac{2119.5}{2} \\ \\ = 1350 - 1059.75 \\ \\ \purple { \boxed{ \therefore \: Area \: of \: shaded \: region = 290.25 \: {cm}^{2}}} \\$

6 0

3 years ago

Write the expresion in simplest form<br> (-11/2x+30)-2(-11/4x-5/2)

qwelly [4]

The simplified expression of (-11/2x+30)-2(-11/4x-5/2) is 35

<h3>How to simplify the expression?</h3>

The expression is given as:

(-11/2x+30)-2(-11/4x-5/2)

Expand the bracket

-11/2x + 30 + 11/2x + 5

Collect the like terms

11/2x -11/2x + 30 + 5

Evaluate the like terms

35

Hence, the simplified expression of (-11/2x+30)-2(-11/4x-5/2) is 35

Read more about expressions at:

brainly.com/question/723406

#SPJ1

4 0

2 years ago

Help me with this...

marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

$333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \: - x = 630 - 801 \\ or \: - x = - 171 \\ x = 171$

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0

4 years ago

Ocean tides can be modeled by a sinusoidal function. Suppose that there is a low and high tide every 12 hours, and that high tid

USPshnik [31]

Answer:

(a) $y=5sin(\frac{\pi}{6}(x+2))+5$

(b) 7.5ft above the low tide.

Step-by-step explanation:

(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:

$y=Asin(B(x-C))+D$

Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.

The amplitude is half the distance between the highest and the lowest tide:

$A=10/2=5ft$

The frequency is related to the period T by:

$B=\frac{2\pi}{T}$

The period is 12 hours, then

$B=\frac{2\pi}{12}=\frac{\pi}{6}$

The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when $sin(B(x-C))=1$ , if $sin(\frac{\pi}{2})=1$ then $B(x-C)$ must be equal to $\frac{\pi}{2}$ when $x=1$ :