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3 years ago

How should the decimal point in 4.003 be moved to determine the product of 100 × 4.003?

Mathematics

1 answer:

tangare [24]3 years ago

7 0

Answer:

Two places to the right.

Step-by-step explanation:

Since theres two zeros in 100, move the decimal two places to the right, making it 400.3.

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Can someone help with this ASAP

LenKa [72]

Answer:

x=7cm.

Step-by-step explanation:

A=1/2(a+b)h.

63=1/2(6+12)x.

63=1/2(18)x.

63=9×x.

63/9=9×x/9.

7cm=x.

x=7cm.

3 0

3 years ago

Show work and answer .............

Lelechka [254]

Answer:

1. 5

2. 5

3. -3

4.y= -1

5. 5

Step-by-step explanation:

1.)x- 2y= -25 ; y=3x

x-2(3x)=(-25)

x-6x=(-25) ; -5x=(-25) ; x=5

2.) 5x+3y =40; x=y

8y=40; y=5

3.) 4x-3y= -27; x=3y

4(3y)-3y= -27; 12y-3y=-27; 9y=-27; y=-3

4.)x-y=9; x=-8y; -8y-y=9; -9y=9; y=-1

5.) 2x+2y=30; y=2x ; 2x+2(2x)=30; 6x=30; x=5

8 0

3 years ago

What is the mean of 37 ?

Sholpan [36]

Is there any more numbers or no

6 0

3 years ago

What is x^2-11x-5=0

mr Goodwill [35]

$x^2-11x-5=0\\a=1,b=-11,c=-5\\D=b^2-4ac=(-11)^2-4\times1\times(-5)=121+20=141\\x_{1,2}=\dfrac{-b\pm\sqrt{D} }{2a} \\x_1=\dfrac{11+\sqrt{141} }{2} \\x_2=\dfrac{11-\sqrt{141} }{2}$

8 0

3 years ago

What are the roots of the polynomial equation x3 - 6x = 3x2 - 8? Use a graphing calculator and a system of equations.

makkiz [27]

Answer:

d : 4, 1, -2.

Step-by-step explanation:

There's no need for a calculator in my opinion because we can use the rational root theorem which states that the equations of this form:

$\sum_{i=0}^{n} a_{i}x^i = 0 = a_{0} + a_{1}x + a_2x^2 + ... + a_nx^n = 0\\ a_i \in \mathbb{Z}$

Have rational roots, than the roots are of the form: $\frac{k \cdot a_o}{p \cdot a_n}, \ where \ k, p \in \mathbb{Z}.$

Rewriting the equation we have:

$x^3 - 3x^2 -6x + 8 = 0\\By \ our\ previous \ claim \ x \in D_8 \ where \ D_n \ the \ set \ of \ divisors \ of \ n.\\D_8 = \{\pm 1, \pm 2, \pm 4, \pm 8\} \\We \ plug \ in \ some \ number from \ D_8.\\Letting \ x = -2;\\(-2)^3 -3(-2)^2 -6(-2) + 8 = 0 \ \ \ Thus \ x+2 \ is \ a \ factor.\\We \ can \ now \ simplify \ the \ eq. \ using \ Polynomial \ Long \ Division.$

$x^3 -3x^2 -6x + 8 = (x+2)(Q(x))\\To \ find \ Q \ we \ divide \ the \ original \ equation \ by \ x + 2; which \ yields:\\Q(x) = x^2 - 5x + 4.\\So \ we \ can \ use \ the \ quadratic \ formula \ to \ find \ the \ roots:\\x = 4,x = 1. \\Thus \ x \in \{-2, 1, 4\}. \ \ These \ are \ all \ the \ roots.$

5 0

4 years ago