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Alex
3 years ago
13

The school store sold a total of 55 shirts for $620. The shirts sold were either blue or gray. If the

Mathematics
1 answer:
PIT_PIT [208]3 years ago
5 0

<u>I used trial and error. It's a great method.</u>

35 blue shirts

20 gray shirts

35*12=420

20*10=200

420+200=620.

Hope this helped! If it didn't or you need further explanation, please tell me so I can do so.

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Brainliest if correct
natali 33 [55]

\left(15^{1/5} \right)^5  = 15^{(5/5)} = 15^1 = 15

6 0
2 years ago
X^2 + bx + 49 is a perfect squad trinomial what is one possible value of b?
ddd [48]

a perfect square trinomial, (x + y)² = x² + 2xy + y²

so, if we have the x of the bx, what is left is the b

the expression would have to be (x + 7)², since we have the 49 and the x²

so, what's left: x² + 14x + 49,

b = 14

<em>hope it helps :)</em>

7 0
3 years ago
Work out the surface area of the solid cuboid.<br> 3cm<br> 4cm<br> 5cm
Zarrin [17]

Answer: 94 cm^3.

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4 0
2 years ago
Read 2 more answers
A landscape supply business $30 to deliver mulch. The cost of the mulch is $23 per cubic yard. Write and solve a linear equation
Andru [333]
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4 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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