MATHEMATICS COLLEGE

Solve for , , and .
-6x+3y-z=5
3x-y+5z=-10
-x+2y+3z=-1

Answers

Answer 1

Answer:

Answer:

x=1, z=-2, y=3

Step-by-step explanation:

Substitution Method

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A particular concentration of a chemical found in polluted water has been found to be lethal to 40% of the crayfish that are exposed to the concentration for 24 hours. 22 crayfish are placed in a tank containing this concentration of chemical in the water. (a) What is the probability that 13 or 17 survive? (b) What is the probability that at least 17 survive? (c) What is the probability that at most 16 survive? (d) What number of crayfish are expected to survive? (e) What is the variance in the number of crayfish that are expected to survive? No actual crayfish were harmed in the making of this question.
What is 32% of 60 as a decimal.
How would I do this?
If x - 10 = -15, then find the value of 5x - 30 *

Is y=3x-3 proportional

Answers

Answer:

i don't think so

Step-by-step explanation:

What is the volume of 22.5 g of metal with a density of 2.81 g/cm

Answers

Answer:V = 72.1 cm³ - 50.0 cm³ = 22.1 cm³

D = mV=99.7g22.1cm³ = 4.51 g/cm³

Step-by-step explanation:

d = mV

m = d×V

V = md

DENSITY

Density is defined as mass per unit volume.

d = mV

Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

Step 1: Calculate the volume

V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³

Step 2: Calculate the density

d = mV = 433g200cm³ = 2.16 g/cm³

MASS

d = mV

We can rearrange this to get the expression for the mass.

m = d×V

Example:

If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?

m = d×V = 500 mL × 1.11g1mL = 555 g

VOLUME

d = mV

We can rearrange this to get the expression for the volume.

V = md

Example:

What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.

Step 1: Convert kilograms to grams.

14.83 kg × 1000g1kg = 14 830 g

Step 2: Calculate the volume.

V = md = 14 830 g × 1cm³19.32g = 767.6 cm³

Why are people with savings hurt by inflation? A. The money is worth more now than it was in the past, leading to a higher tax bill.
B. Inflation discourages people from saving.
C. The money they saved in the past is worth less in the future.
D. Inflation reduces the interest savings accounts pay.

Answers

i took personal finance, sorry if i,m wrong but i think its C

What value should go in the empty boxes to complete the calculation for finding the product of 0.37 * 0.42

Answers

Answer:

0.1554

Step-by-step explanation:

.37x.42=.1554

Final answer:

To find the product of 0.37 and 0.42, multiply the numbers together. The missing digits in the empty boxes should be 5 and 4, respectively.

Explanation:

To find the product of 0.37 and 0.42, you need to multiply the two numbers together. The empty boxes represent the missing digits after the decimal point.

The first number, 0.37, has two decimal places, and the second number, 0.42, has two decimal places as well. When you multiply these two numbers, you need to make sure the total number of decimal places in the product matches the sum of the decimal places in the original numbers.

The product of 0.37 * 0.42 is 0.1554. So, the missing digits in the empty boxes should be 5 and 4, respectively.

Learn more about Multiplication of Decimals here:

brainly.com/question/35552960

#SPJ3

Please helpTrying to help daughter do virtual. And I stink in this stuff!
w(t) = 3t – 1; t = 5

Answers

The answer is W =14/5 I believe

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) π/2 0 3 1 + cos(x) dx, n = 4

Answers

Split up the integration interval into 4 subintervals:

$\left[0,\frac\pi8\right],\left[\frac\pi8,\frac\pi4\right],\left[\frac\pi4,\frac{3\pi}8\right],\left[\frac{3\pi}8,\frac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\frac{i-1}4\left(\frac\pi2-0\right)=\frac{(i-1)\pi}8$

$r_i=\frac i4\left(\frac\pi2-0\right)=\frac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\frac{\ell_i+r_i}2=\frac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\frac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^(\pi/2)\frac3{1+\cos x}\,\mathrm dx\approx\sum_(i=1)^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^(\pi/2)\frac3{1+\cos x}\,\mathrm dx\approx\sum_(i=1)^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)((x-m_i)(x-r_i))/((\ell_i-m_i)(\ell_i-r_i))+f(m)((x-\ell_i)(x-r_i))/((m_i-\ell_i)(m_i-r_i))+f(r_i)((x-\ell_i)(x-m_i))/((r_i-\ell_i)(r_i-m_i))$

so that

$\displaystyle\int_0^(\pi/2)\frac3{1+\cos x}\,\mathrm dx\approx\sum_(i=1)^4\int_(\ell_i)^(r_i)p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_(\ell_i)^(r_i)p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^(\pi/2)\frac3{1+\cos x}\,\mathrm dx\approx\sum_(i=1)^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

Final answer:

The question is asking to approximate the definite integral of 1 + cos(x) from 0 to π/2 using the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule for n=4. These are numerical methods used for approximating integrals by estimating the area under the curve as simpler shapes.

Explanation:

This question asks to use several mathematical rules, specifically the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule, to approximate the given integral with a specified value of n which is 4. The integral given is the function 1 + cos(x) dx from 0 to π/2. Each of these rules are techniques for approximating the definite integral of a function. They work by estimating the region under the graph of the function and above the x-axis as a series of simpler shapes, such as trapezoids or parabolas, and then calculating the area of these shapes. The 'dx' component represents a small change in x, the variable of integration. The cosine function in this integral is a trigonometric function that oscillates between -1 and 1, mapping the unit circle to the x-axis. The exact solution would require calculus, but these numerical methods provide a close approximation.