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3 years ago

If m∠RQS=22x-11 and m∠RQT=12x-8, find m∠TQS

Mathematics

1 answer:

Arada [10]3 years ago

3 0

Answer:

∡TQS = 10x -3

Step-by-step explanation:

∡TQS = ∡RQS - ∡RQT

= 22x - 11 - (12x - 8)

= 10x -3

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Please help if you can!

stealth61 [152]

Given:

The figure of a triangle DEF.

To find:

The coordinate of image of given triangle after reflection over y-axis.

Solution:

The vertices of given triangle are D(1,4), E(4,3), F(2,0).

If a figure is reflected over the y-axis, then the rule of reflection is

$(x,y)\to (-x,y)$

Using this rule, we get

$D(1,4)\to D'(-1,4)$

$E(4,3)\to E'(-4,3)$

$F(2,0)\to F'(-2,0)$

Therefore, the coordinates of image are D'(-1,4), E'(-4,3) and F'(-2,0).

5 0

3 years ago

Find the value of y. Picture below

olga2289 [7]

We have trapezoid with congruent legs.

30x + 11y = 2x + 3

28x +11y = 3

Also we have

30x + 11y = 12

We got 2 equations:

30x + 11y = 12

28x +11y = 3 (*-1) -----> -28x -11y = -3

30x + 11y = 12

-28x -11y = -3

2x = 9

x=9/2

30x + 11y = 12 ------> 30*9/2 +11y =12 ------> 135+11y =12, ---> 11y = - 123,

y= - 123/11

6 0

2 years ago

What is the following product? StartRoot 10 EndRoot times StartRoot 10 EndRoot

pentagon [3]

Answer:

it's 10

Step-by-step explanation:

i searched it up.

5 0

2 years ago

What is the solution to the equation 9 cubed x ≈ 7?

yulyashka [42]

Answer

7/729

Explanation

The question requires to you to find the value of x.

9³x = 7

729 x = 7

x = 7/729

4 0

3 years ago

Find the derivative of

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

6 0

2 years ago