What is the area of Triangle PQR on the grid?

A triangle PQR is shown on a grid. The vertex P is on ordered pair 7 and 6, vertex Q is on ordered pair 1 and 6, and the vertex R is on ordered pair 4 and 2.

5 square units
6 square units
10 square units
12 square units

Answer:

a= [tex]\sqrt{19}[/tex]

Step-by-step explanation:

Pythagorean Theorem: a^2+b^2=c^2

a^2+9^2=10^2

a^2+81=100

a^2=19

a=[tex]\sqrt{19}[/tex]

Answer:

b = 4.4 meters

Answer: False

The variable portions x^2y and xy^2 are slightly different, so that's why we don't have like terms here.

Answer:

False

Step-by-step explanation:

Like terms are those terms where they have the same variable

so here

5x2y variables are = 5x and 2y

7xy2 variables are = 7x and y2

you can observe that x variable is same but y variable is not same as 5x3y have 2y whereas 7xy2 have y2#

so the answer is FALSE

Answer:

7

Step-by-step explanation:

Going from -5 to 2 we get

1) -4

2) -3

3) -2

4) -1

5) 0

6) 1

7) 2

So, in total, there are 7 numbers between -5 and 2

Answer: 79

Concept:

Here, we need to understand the idea of evaluation.  

When encountering questions that gave you an expression with variables, then stated: "If x = a, y = b, z = c" (a, b, c are all constants), this means you should substitute the value given for each variable back to the expression.

Solve:

Given information

10x² - 7x + 10

x = 3

Substitute the value into the expression

= 10 (3)² - 7 (3) + 10

Simplify by multiplication

= 10 (9) - 21 + 10

= 90 - 21 + 10

Simplify by subtraction

= 69 + 10

Simplify by addition

= 79

Hope this helps!! :)

Please let me know if you have any questions

Answer:

92

Step-by-step explanation:

The variable 'x' shows up twice in this expression.  Replace each instance of 'x' with 3:

10(3)^2 - 7(3) + 10  =  10(9) - 21 + 19  =  92

Answer:

i hope it will help

Step-by-step explanation:

I did not get the equation so I solve it with two methods

Answer:

C. 7 units

Step-by-step explanation:

The given parameters are;

The length of the chord of the circle, [tex]\overline{AC}[/tex] = 14 units

The orientation of the radius and the chord = The radius is perpendicular to the chord

We have in ΔAOC, [tex]\overline{AO}[/tex] = [tex]\overline{OC}[/tex] = The radius of the circle

[tex]\overline{OB}[/tex] ≅ [tex]\overline{OB}[/tex]  by reflexive property

The angle at point B = 90° by angle formed by the radius which is perpendiclar to the chord [tex]\overline{AC}[/tex]

ΔAOB and ΔCOB are right triangles (triangles having one 90° angle)

[tex]\overline{AO}[/tex] and [tex]\overline{OC}[/tex] are hypotenuse sides of ΔAOB and ΔCOB respectively and [tex]\overline{OB}[/tex] is a leg to ΔAOB and ΔCOB

Therefore;

ΔAOB ≅ ΔCOB, by Hypotenuse Leg rule of congruency

Therefore;

[tex]\overline{AB}[/tex] ≅ [tex]\overline{BC}[/tex] by Congruent Parts of Congruent Triangles are Congruent, CPCTC

[tex]\overline{AB}[/tex] = [tex]\overline{BC}[/tex] by definition of congruency

[tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] by segment addition postulate

∴ [tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] =  [tex]\overline{AB}[/tex] + [tex]\overline{AB}[/tex] = 2 ×  [tex]\overline{AB}[/tex]

∴  [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex]/2

[tex]\overline{AB}[/tex] = 14/2 = 7

[tex]\overline{AB}[/tex] = 7 units.

Answer:

7 units

Step-by-step explanation:

Answer:

Step-by-step explanation:

The formula for slope is y2-y1/x2-x1 where y2 and x2 are the x and y coordinates from a coordinate pair and y1 and x1 are the coordinates from another coordinate pair. In this case, 2 coordinate pairs are given: (30,75) and (10, 35)  75-35/30-10 would be your slope, or, 40/20, or simplified, 2.

Your slope is 2

Answer:

Step-by-step explanation:

For Part A, use the pythagorean theorem to find the height, which ca be found by finding one length of the leg. Using the imaginary bisector, you can determine one of the legs is 5 cm, and the hypotenuse is 13 cm

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse. Plug in the values to solve for one leg, you get 25+b^2=169

Solve algebraically, b^2= 144, so b=12, which is the height.

Part B

Determine the surface area of each cardboard piece, and add together.

20 × 13 × 2 = 520

1/2 × 10 × 12 × 2 = 120

20 × 10 = 200

So approximately 840 cm of cardboard was used

Answer:

y = 23, -31

Step-by-step explanation

ttyl

Answer:

d no.

Step-by-step explanation:

because in my my personal opinion the function shown on the graph has a smaller rate of change, but a higher starting point

Answer:

x = 2

Step-by-step explanation:

Answer:

27/20

Step-by-step explanation:

[tex]\frac{13.5}{10} = \frac{13.5 * 2}{10 * 2} = \frac{27}{20}[/tex]

Answer:

18[tex]\pi[/tex]

[tex]\frac{80}{360} * 81 \pi[/tex]

Step-by-step explanation:

Answer:

There are 0.454 kg in one pound.

So, in 120 pounds there are 0.454 x 120 kgs.

This is equal to 54.48, and the answer is 54.48 kg.

Let me know if this helps!

Answer:

When you plot the graph, you may see a line or a curve. This presents the trend of the relationship of the two quantities when the independent variable changes.

Graphs visually depict patterns and trends, while equations provide precise mathematical descriptions, both revealing valuable information about the relationship between two quantities.

Graphs visually represent data and patterns, allowing us to identify the nature of the relationship between two quantities. The slope and intercepts of a graph provide information about the rate of change and the values when one quantity is zero.

Equations, on the other hand, offer precise mathematical descriptions of the relationship. The form of the equation reveals the nature of the relationship, while coefficients and constants provide specific information. Solving equations yields solutions that satisfy the relationship.

Both graphs and equations work together to provide a comprehensive understanding of the relationship between two quantities, with graphs offering visual insights and equations providing precise mathematical information.

Learn more about equations here;

https://brainly.com/question/4803354

#SPJ6

Answer:

$420

.8 x = 336

x = 336/.8

X=$420

Step-by-step explanation:

Answer:

0.63 ÷ 0.3 = 63 x 10^-2 ÷ 3 x 10^-1 = 63 / 3  x 10^(-2--1) = 21 x 10^-1 = 2.1

Answer:

2.1.

Step-by-step explanation:

63/3 = 21

0.63 = 63 / 100

and 0.3 = 3 / 10

so 0.63/0.3 = 63 /100 / 3/10 = 63/100 * 10/3

= 63/3 / 10

So the answer is 21 / 10 = 2.1.

Answer:

1-10, -3, 2, 7

im sorry if i give a wrong answer

Answer

25mm of rain fell each second

Step-by-step explanation:

we know that 500 mm fell in 20 minutes

so, we have to divide 500 by 20 giving us the amount of rain that fell each minute:

500/20 = 25

therefore, 25mm of rain fell each minute

hope this helped:)

Answer:

25 mm per minute

Step-by-step explanation:

Take the amount of rain and divide by the number of minutes

500/20

25 mm per minute

no cheating thanks <3

Answer:

where are the numbers

Step-by-step explanation:

Answer:

A.1/3 B.5/8 C.1/25

Step-by-step explanation:

just simplify it down to the lowest

Answer:

Ä. 16/24 as reduced fraction = 2/3

B. 15/24 as reduced fraction= 5/8

Answer:

The property showed in the number 21 is the distributive property

Answer:

a, a, d

Step-by-step explanation:

44

The area (A) of a triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the perpendicular height )

Here b = 10 and h = 9 , then

A = [tex]\frac{1}{2}[/tex] × 10 × 9 = 5 × 9 = 45 in² → a

45

The area (A) of a parallelogram is

A = bh ( b is the base and h the perpendicular height )

Here b = 2 and h = 4 , then

A = 2 × 4 = 8 m² → a

46

A = bh ( with b = 4 and h = 10 )

A = 4 × 10 = 40 m² → d

Answer:

Step-by-step explanation:

[tex]\displaystyle \ \Large \boldsymbol{} \frac{a-\sqrt{a} }{\sqrt{a}-1 } -\frac{\sqrt{a}+1 }{a+\sqrt{a} } :\frac{\sqrt{a}+1 }{a} = \\\\\\\frac{\sqrt{a}(\sqrt{a} -1 ) }{(\sqrt{a}-1) } -\frac{\sqrt{a}+1 }{\sqrt{a}(\sqrt{a}+1 )}\cdot \frac{\sqrt{a}\cdot \sqrt{a} }{\sqrt{a}+1 } = \\\\\\\sqrt{a} -\frac{\sqrt{a} }{1+\sqrt{a} } =\frac{a+\sqrt{a}-\sqrt{a} }{1+\sqrt{a} } = \\\\\\\frac{a}{\sqrt{a}+1 } \cdot \frac{\sqrt{a}-1 }{\sqrt{a}-1} } =\boxed{\frac{a\sqrt{a} -a}{a-1} }[/tex]

Answer:

[tex]a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}[/tex]

By expanding the expression, we get;

[tex]\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}[/tex]

Collecting like terms gives;

[tex]\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

(b) The given expression is presented as follows;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4[/tex]

Therefore, we get;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}[/tex]

Collecting like terms gives;

[tex]x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )[/tex]

[tex]x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]