draw the graph of the line: with a slope of -1/2 and that passes through the origin

The triangle is missing and so i have attached it.

Answer:

x = 93°

Step-by-step explanation:

From the triangle attached, we can say that;

<SQP + <SQR = 180°

This is because sum of angles on a straight line equals 180°.

Secondly, we know that sum of angles in a triangle also equals 180°.

Thus;

<SQR + <QSR + <SRQ = 180

From the attached triangle, we see that;

<QSR = 35°

<SRQ = 58°

Thus;

<SQR + 35° + 58° = 180°

<SQR + 93° = 180°

<SQR = 180° - 93°

<SQR = 87°

From earlier on, we saw that;

<SQP + <SQR = 180°

Plugging in <SQR = 87°, we have;

<SQP + 87° = 180°

<SQP = 180° - 87°

<SQP = 93°

We are told in the question that <SQP is denoted by x.

Thus;

x = 93°

Answer:

The value of x is answer D: 93

Answer:

2 × 2 × 3 × 5

Step-by-step explanation:

Given that,

The number = 60

To find,

Factors of 60 greater than 1 = ?

Procedure:

As we know,

Any of various numbers multiplied together to form a whole.

To find the factors of a number, we will have to do its prime factorization.

So,

The prime factorization of 60:

1 * 2 * 2 * 3 * 5 = 60

Since the factors greater than 1 are asked, the factors would be;

2 * 2 * 3 * 5

Thus, 2 * 2 * 3 * 5 is the correct answer.

Replace x with 3 and solve:

10 x 5^3

Simplify:

10 x 125 = 1250

Answer: y = 1250

(3 , 1250)

I got it right on my test.

Answer:

2 is the answer . the explanation is in the attachment .

Answer:

This would be a .04 or 4% decay.....

for every "time unit" (x in this case) you will be multiplying

the amount by .96 ... in other words if you started with one dollar

the results would be 96 cents... after two "time" steps you would have

only 92 cents (.96 *.96)

Step-by-step explanation:

Answer:

11 terms.

Step-by-step explanation:

We are given the arithmetic sequence:

17, 13, 9, ...

And we want to find the number of terms required such that the sum is -33.

Recall that the sum of an arithmetic series is given by:

[tex]\displaystyle S = \frac{k}{2}\left( a + x_k\right)[/tex]

Where k is the number of terms, a is the first term, and x_k is the last term.

The desired sum is -33. The first term is 17 as well. Thus:

[tex]\displaystyle (-33) = \frac{k}{2} \left( (17) +x_k\right)[/tex]

Simplify:

[tex]-66 = k(17 + x_k)[/tex]

We can write a direct formula to find the last term x_k. The direct formula of an arithmetic sequence has the form:

[tex]x_ n = a + d(n-1)[/tex]

Where a is the initial term and d is the common difference.

The initial term is 17 and the common difference is -4. Hence:

[tex]\displaystyle x_n = 17 - 4(n-1)[/tex]

Then the last term is given by:

[tex]x_k = 17 - 4(k-1)[/tex]

Substitute:

[tex]\displaystyle -66 = k\left( 17 + \left( 17 - 4(k-1)\right)\right)[/tex]

Solve for k:

[tex]\displaystyle \begin{aligned} -66 &= k(17 + (17 - 4k + 4)) \\ -66 &= k(38 -4k) \\ -66 &= -4k^2 + 38k \\ 4k^2 -38k -66 &= 0 \\ 2k^2 - 19k -33 &= 0 \\ (k-11)(2k+3) &= 0 \\ k-11&= 0 \text{ or } 2k+3 = 0 \\ \\ k &= 11 \text{ or } k = -\frac{3}{2}\end{aligned}[/tex]

Since we cannot have a negative amount of terms, we can ignore the second solution.

Therefore, the given sequence must have 11 terms such that it sums to -33.

Answer:

Here is 2 methods

Step-by-step explanation:

1) we use excel to find n=11 for lasy students

2) mathematical method

[tex]u_1=17\\u_2=13=17+(2-1)*(-4)\\u_3=9=17+(3-1)*(-4)\\\\\\\boxed{u_n=17+(n-1)*(-4)}\\\\\\\displaystyle s_n=\sum_{i=1}^nu_i\\=\sum_{i=1}^n(17+(i-1)*(-4))\\\\\\=(\sum_{i=1}^n 17) + (-4)*\sum_{i=1}^n (i) +4*\sum_{i=1}^n (1)\\\\\\=17*n+4*n-4*\frac{n*(n+1)}{2} \\\\\\=21n-2n^2-2n\\\\\\=-2n^2+19n\\\\=-33\\\\\\\Longrightarrow\ 2n^2-19n-33=0[/tex]

[tex]\Delta=19^2+4*2*33=625=25^2\\\\n=\dfrac{19-25}{4} =-1.5\ (excluded)\ or\ n=\dfrac{19+25}{4}=11\\\\[/tex]

Answer:

Step-by-step explanation:

If you plot those coordinates in your calculator in the stat plot function under "stat", you will see that the dots are almost but not quite in a straight line going from upper left to lower right. This indicates a strong negative linear association, third choice down.

Answer:

D. Nonlinear Negative Association

Step-by-step explanation:

Answer:

emily fills 3 racks

3 basketballs are Left!

Step-by-step explanation:

15/4 = 3.3

Answer:

A =147 cm^2

Step-by-step explanation:

A = pi r^2

The radius is 7 and let pi = 3

A = 3*7^2

A = 3*49

A =147

Step-by-step explanation:

[tex]s = \pi \times {r}^{2} = \pi \times {6}^{2} = 36\pi[/tex]

[tex]h = 18 \times \sin(60) = 9 \sqrt{3} [/tex]

[tex]v = s \times h = 36\pi \times 9 \sqrt{3} = 324 \sqrt{3} \pi[/tex]

Answer:

VERTICALLY OPP ANGLES

Step-by-step explanation:

9514 1404 393

Answer:

  -4/3

Step-by-step explanation:

Quadratic ax² +bx +c can be written in factored form as ...

  a(x -p)(x -q)

for zeros p and q. The expanded form of this is ...

  ax² -a(p+q)x +apq

Then the ratio of the constant term to the leading coefficient is ...

  c/a = (apq)/a = pq . . . . the product of the zeros

For your quadratic, the ratio c/a is -4/3, the product of the zeros.

_____

Additional comment

You will notice that the sum of zeros is ...

  -b/a = -(-a(p+q))/a = p+q

Answer:

[tex] \green{ \boxed{ \bf \: product \: of \: the \: zeros \: = - \frac{4}{3} }}[/tex]

Step-by-step explanation:

We know that,

[tex] \sf \: if \: \alpha \: and \: \beta \: \: are \: the \: zeroes \: of \: the \: \\ \sf \: polynomial \: \: \: \pink{a {x}^{2} + bx + c }\: \: \: \: then \\ \\ \small{ \sf \: product \: of \: zeroes \: \: \: \alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } } \\ \\ \sf \implies \: \pink{ \boxed{\alpha \beta = \frac{c}{a} }}[/tex]

Given that, the polynomial is :

[tex] \bf \: 3 {x}^{2} - 2x - 4[/tex]

so,

[tex] \sf \: so \: product \: of \: zeroes \: \: = \frac{ - 4}{3} = - \frac{4}{3} [/tex]

Answer: A x> -2

Step-by-step explanation:

Answer: The range is  [tex]\{y \in \mathbb{R}\ | \ y \le -6\}[/tex]

Explanation:

The parabola opens down, forming a "frowny face" in a way (just without the eyes). Or you can think of it as a hill or mountain. This means that the vertex (-3,-6) is at the top of that mountain. It's the highest point of that parabola.

The range is the set of all possible y values. We see that y = -6 is the largest it can get. So y = -6 or y is smaller than this. We would then write [tex]y \le -6[/tex] to describe all the possible y values.

Therefore, the range is [tex]\{y \in \mathbb{R}\ | \ y \le -6\}[/tex]

This translates to "y is a real number such that y is -6 or smaller".

So the second answer you wrote is close, but you forgot the "or equal to" portion of the inequality sign.

See below for a visual example of what's going on.

Answer:

answer is 21..............

Explanation:

p(x) = x⁴ + 2x³- 3x² + x - 1

Factor of p(x)

x-2=0

x=2

Then by using synthetic division

Answer:

Area of the slice of pie = 22.09 ft²

Step-by-step explanation:

Area of the slice of pie = Area of the sector of the circle with the central angle 45°

Area of the sector = [tex]\frac{\theta}{360^{\circ}}(\pi r^{2} )[/tex] [Here, r = radius of the circle]

                              = [tex]\frac{45^{\circ}}{360^{\circ}}(\pi )(\frac{15}{2})^2[/tex]

                              = 22.09 ft²

Area of the slice of pie = 22.09 ft²

Answer:

22.08ft^2

Step-by-step explanation:

A = πr^2(x/360) d = 15

Since r is half of diameter this means that r = 15/2 =7.5

so Lets use the Area of Sector formua

A =3.14(7.5)^2 (45/360)

A =3.14(56.25) (45/360)

A = 176.625 (45/360)

A = 176.625 (0.125)

A = 22.078125

rounded to the nearest 10th would make it 22.08

Answer:

2.25 km to go

Step-by-step explanation:

In order to get this answer, you have to figure out how many km 10% is, 0.1125. Then multiply that by the remaining 20% because he already finished 80% of the trail. So, 0.1125 x 20 = 2.25.

Hope this helps! :)

Answer:

[tex]{ \tt{ \tan {}^{4} \theta + { \sec }^{2} \theta }} \\ { \tt{ = ({ \tan }^{2} \theta ){}^{2} + { \sec }^{2} \theta }} \\ = { \tt{ {-(1 - { \sec }^{2} \theta) }^{2} + { \sec }^{2} \theta }} \\ { \tt{ = -(1 - 2 { \sec }^{2} \theta + { \sec }^{4} \theta) + { \sec}^{2} \theta}} \\ { \tt{ = -(1 - { \sec }^{2} \theta) + { \sec }^{4} \theta}} \\ { \tt{ = -{ \tan}^{2} \theta + { \sec }^{4} \theta }} \\ = { \tt{ { \sec}^{4} \theta - { \tan }^{2} \theta}} \\ { \bf{hence \: proved}}[/tex]

Use distributive law

[tex]\boxed{\sf a(b+c)=ab+ac}[/tex]

Now

[tex]\\ \sf\longmapsto 2x(x-y)+3y(x-y)[/tex]

[tex]\\ \sf\longmapsto 2x^2-2xy+3xy-3y^2[/tex]

[tex]\\ \sf\longmapsto 2x^2-3y^2-2xy+3x^2[/tex]

[tex]\\ \sf\longmapsto 2x^2-3y^2+xy[/tex]

Taking common

Answer: 2x (x-y) + 3y (x-y)

             = ( x-y ) ( 2x-3y )

Answer:

I think it's 43 and 43 degrees. I just subtracted 180-94, got 86, and divided it so yea.

Answer:

Step-by-step explanation:

We can see that there are roots at (-2,0) and (-1,0)

also, the root at (-2,0) should bounce right off

and the root at (-1,0) should go through

With all that being said it has to be B

Answer:

r = 13.33

Step-by-step explanation:

d = k*r*t

Where,

k = constant of proportionality

d = 110 when r = 55 and t = 2

d = k*r*t

110 = k * 55 * 2

110 = 110k

k = 110/110

k = 1

Find r when d = 40 and t = 3

d = k*r*t

40 = 1 * r * 3

40 = 3r

r = 40/3

= 13.333333333333

Approximately,

r = 13.33

Answer:

W = 7 - x

Step-by-step explanation:

The perimeter is P= 2L + 2×W , where L is the length and W is the width.

If L = (x+2) , replacing L with the expression x+2 we have

P= 2×(X+2) + 2W ⇔ 18 = 2x + 4 + 2W ⇔ 2W =18 - 2x - 4 ⇔ 2W = 14 - 2x

⇔ W = 7 - x

Answer:

X+y=7

Step-by-step explanation:

i  remember  doing something like this but mines had the word onion rings .

Answer:

The rea of walk way is 32 ft^2.

Step-by-step explanation:

width, w = 8 feet

length, L = 10 feet

width of walkway, d = 2 feet

length of outer, L' = 10 + 2 + 2 = 14 feet

Area of outer, A' = L' x w = 14 x 8 = 112 ft^2

Area of inner, A = L x w = 10 x 8 = 80 ft^2

The area of walkway = A' - A = 112 - 80 = 32 ft^2

The rea of walk way is 32 ft^2.

If Forces are acting on opposite direction

[tex]\\ \rm\hookrightarrow F_{net}=F_2-F_1[/tex]

[tex]\\ \rm\hookrightarrow F_{net}=10-6[/tex]

[tex]\\ \rm\hookrightarrow F_{net}=4N[/tex]

If both acting on same direction

[tex]\\ \rm\hookrightarrow F_{net}=F_1+F_2[/tex]

[tex]\\ \rm\hookrightarrow F_{net}=10+6[/tex]

[tex]\\ \rm\hookrightarrow F_{net}=16N[/tex]

Hence

[tex]\boxed{\sf 4N\leqslant F_{net}\leqslant 16N}[/tex]