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2 years ago

Multiply. Just multiply the decimals. Please?

Mathematics

2 answers:

Alex17521 [72]2 years ago

5 0

1) 0.009

2) 0.294

3) 0.0612

4) 33.62

5) 26.46

6) 1.824

Finger [1]2 years ago

4 0

1) 0.009

2) 0.294

3) 0.0612

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Question: The graph of a second-order polynomial is shown below, and the intercepts with the axes are marked. Explain how you ca

notka56 [123]

Answer:

Step-by-step explanation:

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2 years ago

Maths functions question!!

Marina86 [1]

Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

8 0

4 months ago

Read 2 more answers

What are 2 points that satisfy the equation x+y = -5

777dan777 [17]

Answer: (0,-5)(1,-6) is the answers

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2 years ago

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$