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3 years ago

Write a number that has a zero in the tens place. Draw a quick picture for your number

2 answers:

7 0

102. I can't draw on brainly, but u can make a table of ones, tens, hundreds. Then under the hundreds, draw one circle to represent 100. Leave the ten column blank, and then draw two circles under the ones, to represent the 2 ones.

natali 33 [55]3 years ago

4 0

400 hope that helped

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The table shows the total cost of different numbers of tickets decided whether it makes sense to use a constant rate to describe

GalinKa [24]

34.50/6 = 5.75

44/8 = 5.5

52.5/10 = 5.25

As the number of tickets increase, the cost per ticket decreases.

So it does not make sense to use a constant rate.

5 0

1 year ago

Read 2 more answers

Help me please!!

bixtya [17]

Answer:

The measurement of angle B is 135°

The measurement of angle A is 60°

The measurement of angle C is 75°

Step-by-step explanation:

Angle A is the same value of 60, Angle B is pretty much just 180 - 45, and so is angle C.

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3 years ago

State the leading coefficient of the polynomial: f(x) = 6x^5 + 9x^4 + 2x^3 + 3x^2 - 4x -6

dlinn [17]

Answer:

The leading coefficient of the polynomial is 5 .

Step-by-step explanation:

f(x) = 6x^5 + 9x^4 + 2x^3 + 3x^2 - 4x -6

In this polynomial , the highest exponential is the leading coefficient .

So there are 5 coefficients ,

5 , 4 , 3 , 2 , and x

so above , the highest one is 5 .

4 0

3 years ago

Layla has 5.49 pounds of wood she divides it into 7 so basically 5.49÷7

Nuetrik [128]

Answer:

0.7842857142857143

Step-by-step explanation: i would right an explanation but i cant put it all so you will just have to trust me

7 0

3 years ago

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f(x) = x2 – 2x + 2 f(x) = x3 – x

konstantin123 [22]

Answer:

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

Step-by-step explanation:

we know that

The conjugate root theorem states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial

In this problem we have that

The polynomial has roots 1 and (1+i)

by the conjugate root theorem

(1-i) is also a root of the polynomial

therefore

The lowest degree of the polynomial is 3

$f(x)=a(x-1)(x-(1+i))(x-(1-i))$

Remember that

The leading coefficient is 1

a=1

$f(x)=(x-1)(x-(1+i))(x-(1-i))\\\\f(x)=(x-1)[x^{2} -(1-i)x-(1+i)x+(1-i^2)]\\\\f(x)=(x-1)[x^{2} -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^{2} -2x+2]\\\\f(x)=x^{3}-2x^{2} +2x-x^{2} +2x-2\\\\f(x)=x^{3}-3x^{2} +4x-2$

5 0

3 years ago

Read 2 more answers