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3 years ago

Solve the equation. Enter the answer as an equation showing the value of the

Mathematics

2 answers:

Mice21 [21]3 years ago

8 0

<h3>Question:-</h3>

Solve the equation. Enter the answer as an equation showing the value of the variable (like s = -6, or w=1/2). Use a slash mark (/) as the fraction bar if necessary

X/10 = 140

<h3>Answer:-</h3>

x = 1400

<h3>Explanation:-</h3>

$\rightarrow \: \frac{x}{10} = 140$

Multiply 10 on each side,

$\rightarrow \: 10 \times \frac{x}{10} = 140 \times 10$

$\rightarrow \: 1 × x= 1400$

$\tt \red \rightarrow \: x = 1400$

marin [14]3 years ago

3 0

Answer:

x = 1400

Step-by-step explanation:

→ x/10 = 140

Multiply 10 on both sides.

→ x/10 × 10 = 140 × 10

→ x × 1 = 1400

→ x = 1400

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A person standing close to the edge on top of a 72-foot building throws a ball vertically upward. The quadratic function h(t)=−1

aleksklad [387]

The quadratic function h(t) = −16t² + 84t + 72, shows that the person throws the ball from on top of a 72 feet building at a velocity of 84 feet/second.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

The quadratic function h(t) = −16t² + 84t + 72, shows that the person throws the ball from on top of a 72 feet building at a velocity of 84 feet/second.

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2 years ago

The table shows the location of four cities relative to sea level.

ycow [4]

Answer:its citc (C)

Step-by-step explanation:

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3 years ago

Show work on how you converted them. (also marking brainliest) <3

Dmitry_Shevchenko [17]

Answer:

1. 5700mg

2. 1000cm

3. 0.0453m

4. 37.5 kg/L

Step-by-step explanation:

1. 1gram (g) =1000milligrams(mg)

5.70×1000=5700mg

2. 1 m= 100cm

10×100=1000cm

3. 1000mm=1m

45.3/1000=0.0453m

4. 1000g=1kg

37.5/1000=0.0375

1000ml=1L

1ml= 1/1000= 0.001L

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8 0

3 years ago

I need to find the surface area and volume of all three figures. If you could provide what equations you used to I will be grate

Margarita [4]

Answer:

Part 1) Sphere The surface area is equal to $SA=196\pi\ m^{2}$ and the volume is equal to $V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) Cone The surface area is equal to $SA=(16+4\sqrt{65})\pi\ units^{2}$ and the volume is equal to $V=\frac{112}{3}\pi\ units^{3}$

Part 3) Triangular Prism The surface area is equal to $SA=51.57\ mm^{2}$ and the volume is equal to $V=17.388\ mm^{3}$

Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

$SA=4\pi r^{2}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$SA=4\pi (7)^{2}$

$SA=196\pi\ m^{2}$

b) Find the volume

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (7)^{3}$

$V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) The figure is a cone

a) Find the surface area

The surface area of a cone is equal to

$SA=\pi r^{2} +\pi rl$

we have

$r=4\ units$

$h=7\ units$

Applying Pythagoras Theorem find the value of l (slant height)

$l^{2}=r^{2} +h^{2}$

substitute the values

$l^{2}=4^{2} +7^{2}$

$l^{2}=65$

$l=\sqrt{65}\ units$

$SA=\pi (4)^{2} +\pi (4)(\sqrt{65})$

$SA=16\pi +4\sqrt{65}\pi$

$SA=(16+4\sqrt{65})\pi\ units^{2}$

b) Find the volume

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=4\ units$

$h=7\ units$

substitute

$V=\frac{1}{3}\pi (4)^{2}(7)$

$V=\frac{112}{3}\pi\ units^{3}$

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

$SA=2B+PL$

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

Find the area of the base B

$B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}$

Find the perimeter of the base P

$P=2.7*3=8.1\ mm$

we have

$L=5.6\ mm$

substitute the values

$SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}$

b) Find the volume

The volume of the triangular prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$B=3.105\ mm^{2}$

$L=5.6\ mm$

substitute

$V=(3.105)(5.6)=17.388\ mm^{3}$

5 0

3 years ago

A) Describe the congruence shortcut (postulate/theorem) that could be used to prove that ΔABC≅ΔDEF

bixtya [17]

Answer:

A. By the Side-Side-Side (SSS) Congruence Theorem, ΔABC ≅ ΔDEF.

B. $\overline{AB} \cong \overline{DE}$

$\overline{BC} \cong \overline{EF}$

$\overline{AC} \cong \overline{DF}$

Step-by-step explanation:

a. From the diagram given, it shows that all three sides of ∆ABC are congruent to all the three corresponding sides of ∆DEF.

Therefore, by the Side-Side-Side (SSS) Congruence Theorem, ΔABC ≅ ΔDEF.

b. The sides that are congruent in the ∆s given that applies to the SSS Congruence Theorem are:

$\overline{AB} \cong \overline{DE}$

$\overline{BC} \cong \overline{EF}$

$\overline{AC} \cong \overline{DF}$

6 0

3 years ago