What is the answer to 9+3/4x=7/8x-10

Which shapes can be drawn with two opposite angles equal to 105°

sertanlavr [38]

Answer:

see the explanation

Step-by-step explanation:

we know that

A shape with two opposite angles equal to 105° could be a quadrilateral, a parallelogram, a rhombus or a trapezoid

Because

A quadrilateral: A quadrilateral is a four-sided polygon. The sum of the interior angles in any quadrilateral must be equal to 360 degrees

so

If the quadrilateral have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to

$360^o-2(105^o)=150^o$

A parallelogram: A Parallelogram is a flat shape with opposite sides parallel and equal in length. Opposite angles are congruent and consecutive angles are supplementary

so

If the parallelogram have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to

$180^o-105^o=75^o$

A rhombus: A Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond. All sides have equal length. Opposite sides are parallel. Opposite angles are congruent and consecutive angles are supplementary

so

If the Rhombus have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to

$180^o-105^o=75^o$

A trapezoid: A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel

so

If the trapezoid have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to

$360^o-2(105^o)=150^o$

4 0

3 years ago

1. What is the solution? y 3 -3-2-1 2 3x O (1,2) O (1,-1) O 1 02

Rama09 [41]

9514 1404 393

Answer:

(d) Infinitely Many Solutions

Step-by-step explanation:

Each point of intersection between the lines is a solution. When the lines lie on top of each other, there are infinitely many points of intersection, hence ...

Infinitely Many Solutions

4 0

3 years ago

If y = kx, what is the value of k if y = 20 and x = 0.4? A) 5 B) 8 C) 9.6 D) 50

mariarad [96]

Answer:

D) 50

20/0.4=50

Step-by-step explanation:

7 0

3 years ago

Help please!!! I dont understand these questions currently attaching photos dont delete

Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
y⁸/x¹⁰
m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

__

4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$

3 0

3 years ago

Please help with geometry. 20 points, and another question worth 98 on my profile!

vovikov84 [41]

Answer:

Part 1) m∠1 =(1/2)[arc SP+arc QR]

Part 2) $PR^{2} =PS*PT$

Part 3) PQ=PR

Part 4) m∠QPT=(1/2)[arc QT-arc QS]

Step-by-step explanation:

Part 1)

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

we have

m∠1 -----> is the inner angle

The arcs that comprise it and its opposite are arc SP and arc QR

so

m∠1 =(1/2)[arc SP+arc QR]

Part 2)

we know that

The Intersecting Secant-Tangent Theorem, states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

so

In this problem we have that

$PR^{2} =PS*PT$

Part 3)

we know that

The Tangent-Tangent Theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments

so

In this problem

PQ=PR

Part 4)

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

In this problem

m∠QPT -----> is the outer angle

The arcs that it encompasses are arc QT and arc QS

therefore

m∠QPT=(1/2)[arc QT-arc QS]

4 0

3 years ago