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3 years ago

If an object has a blade, then it is a kitchen knife.

Mathematics

2 answers:

choli [55]3 years ago

7 0

B a kitchen knife? I don’t understand

cluponka [151]3 years ago

7 0

Answer:

a) a pair of scissors

explanation:

The blades of the scissors are a pair of wedges in a cross shape which pivot around a fulcrum. The attached handles are levers.

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Dean is ready to buy a new car. He has $24,000 to spend. The sales tax where he lives is 6%. He put his research on car prices i

mixer [17]

The most expensive car Dean can buy is the 4-door hatchback.

22500 x .06 = 13501350 + 22500 = 23850

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3 years ago

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The mall is 4.72 kilometers from Julie’s house and 1.83 kilometers from Taylor’s house. How much farther does Julie live from th

blondinia [14]

Julies lives 2.89 kilometers more from the mall than Taylor.

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3 years ago

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Evaluate f(x)=4x+1 when x=−1, x=0, and x=3.

larisa86 [58]

Answer:

when

Step-by-step explanation:

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3 years ago

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How much more is needed to make $10.00 4 dollars, 12 dimes, and 10 pennies

amid [387]

Answer:

4 dollars

7 dimes or $4.70.

Step-by-step explanation:

The 4 dollars, 12 dimes, and 10 pennies added up together is $5.30. You will subtract that from $10.00 and you will get $4.70. That will be 4 dollars and 7 dimes left that is needed, or $4.70.

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2 years ago

What are the possible rational roots of the polynomial equation? 0=2x7+3x5−9x2+6

RoseWind [281]

Answer: $\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

$a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0$

Where:

- The coefficients are integers.

- $a_n$ is the leading coeffcient ( $a_n\neq 0$ )

- $a_0$ is the constant term $a_0\neq 0$

Every rational root of the polynomial is in the form:

$\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}$

For the case of the given polynomial:

$2x^7+3x^5-9x^2+6=0$

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

Then, by Rational Roots Test we get the possible rational roots of this polynomial:

$\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

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3 years ago