A block of gold has a mass of 9,660 kilograms has a volume of 0.5 cubic meters. What is the density of gold

Sooo ummm help... 15-p=p-3

bearhunter [10]

Answer:

p = 9

Step-by-step explanation:

6 0

2 years ago

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Evaluate \dfrac j4 4 j ? start fraction, j, divided by, 4, end fraction when j=12j=12j, equals, 12.

cricket20 [7]

We need to evaluate the value of fraction $\frac{j}{4}$ .

We also given j=12.

In order to evaluate the value of fraction $\frac{j}{4}$ , we need to plug j=12.

Plugging j=12, we get

$\frac{j}{4}= \frac{12}{4}$

We have 12 in numerator and 4 in denominator.

We always divide top number by bottom number.

So, we need to divide 12 by 4.

On dividing 12 by 4 we get 3.

<h3>Therefore, $\frac{j}{4} = 3.$ </h3>

3 0

3 years ago

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0.05x =0.5 can someone please helppppp!!!!!!!!!!

Tcecarenko [31]

Answer:

x equals 10

Step-by-step explanation:

you are moving the 5 from the hundredth value to the tenth, meaning it is becoming a higher number. for this one, just multiply by 10.

4 0

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

9 months ago

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Need urgent help doing this question!

xxMikexx [17]

⟘ = 90°

The intersect point is at the N not P, so the answer is m∠BNP = 90°

8 0

3 years ago