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2 years ago

The middle school band has a boy to girl ratio of 4:6.

Mathematics

1 answer:

Elan Coil [88]2 years ago

7 0

4x+ 6x= 45
10x= 45
x= 4,5
girl= 6*4,5= 27

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Given: m∠A = m∠C = 90° AB ║ DC Prove: ∆ABD ≅ ∆CDB 40 points

leonid [27]

Answer:

ABD = CDB

AB is parallel to DC

Step-by-step explanation:

angle a and c are right angles so they are equal and angle b and d are equal because they have a line that touch both letters

AB and DC are parellel because they are the same looking lines just on oppisite sides

7 0

2 years ago

−4p+9=−5 Help a sister out plz

Lorico [155]

Answer: p=7/2=3.5

Step-by-step explanation:

Given

-4p+9=-5

Subtract 9 on both sides

-4p+9-9=-5-9

-4p=-14

Divide -4 on both sides

-4p/-4=-14/-4

p=7/2=3.5

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4 0

2 years ago

Read 2 more answers

Determine the unit rate then determine the better buy Option A: a 4 pack of 3.2-ounce servings of apple sauce for $2.50 or Optio

Greeley [361]

Answer:The better buy Option is A

Step-by-step explanation:

4 0

2 years ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

2 years ago

Evaluate the integral using integration by parts with the indicated choices of u and dv. (Use C for the constant of integration.

trasher [3.6K]

Answer:

$\frac{xe^{7x}}{7} + \frac{e^{7x}}{49}$

Step-by-step explanation:

Given the integral equation

$\int\limits{xe^{7x}} \, dx \\$

According to integration by part;

$\int\limits {u} \, dv = uv + \int\limits {v} \, du$

u = x, dv = e^7x

du/dx = 1

du = dx

$v = \int\limits {e^{7x}} \, dx \\v = e^7x/7$

Substitute the given values into the formula;

$\int\limits {xe^{7x}} \, dx = x(e^{7x}/7) + \int\limits ({e^{7x}/7}) \, dx\\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{7*7} \\\int\limits {xe^{7x}} \, dx = \frac{xe^{7x}}{7} + \frac{e^{7x}}{49}$

3 0

2 years ago