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Juli2301 [7.4K]
2 years ago
12

Determine the equation of a line with a slope of 2 and passes through (-2,1).

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
8 0
sure i will help u.
You might be interested in
Help me :}
nadezda [96]
6the answer will be b 

7 0
3 years ago
4.5w = 5.1w - 30<br><br> Please help!!!
Lelu [443]
4.5w = 5.1w - 30

4.5w - 5.1w = -30

-0.6w =-30

Divide both sides by -0.6

-0.6w/-0.6 = -30/-0.6

<span>w = 50

</span>I hope this helps.
6 0
3 years ago
Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks. Quincy and his mo
Vikentia [17]

Step-by-step explanation:

rc = planted rows of carrots

rt = planted rows of tomatoes

h = number of hours

rc = 4h + 1 (starting with 1 already finished row there will be 4 additional rows every hour)

rt = 3h + 8 (staying with 8 already finished rows there will be 3 additional rows every hour).

when will they have planned the same number of rows ?

when rc = rt, of course.

so,

4h + 1 = 3h + 8

h = 7

4×7 + 1 = 28 + 1 = 29 rows

after 7 hours Quincy and his mom will each have planted 29 rows of vegetables.

8 0
2 years ago
7 and 12 Find the measure of angle x. Round your answer to the nearest hundredth. (please type the numerical answer only) (30 Po
goldfiish [28.3K]

Answer:

Angle Y = 35.69 degrees

Step-by-step explanation:

We can use some simple trigonometry to help us out. Here we know that side 7 is the opposite side relative to angle y, and  12 is the hypotenuse.

We think what trig ratio involves comparing opposite and hypotenuse. Sine is opposite over hypotenuse and we can set up a equation to solve.

Sine (Angle Y) = Opposite Side / Hypotenuse

Sine (Angle Y) = 7/12

(Angle Y) = Sine inverse (7/12)

Angle Y = 35.69

8 0
2 years ago
Write a polynomial of degree 5 with zero x=0,i square root 7, -2i
professor190 [17]

Answer:

P(x)=x^5+11x^3+28x

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

7 0
3 years ago
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