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3 years ago

Approximately how many times greater is 4×107 than 8×103 ?

Mathematics

2 answers:

bixtya [17]3 years ago

7 0

40000000

8000

40000000 / 8000 = 5000

Answer is C. 5000

guajiro [1.7K]3 years ago

4 0

The answer is C. 5000

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Find the equation of this line

Dima020 [189]

10 -3.5 and 10 -8.5 is the answer I think

7 0

2 years ago

Whats the density of an object with the length of 5.0 cm a height of 3.0 and width of 15.0 cm and a mass of 24 grams

Sphinxa [80]

Answer:

The density of the object is $0.1067$ grams per cubic centimeter. (Rounded off to 4th decimal)

Step-by-step explanation:

Density = [Mass/Volume]

Here we have been given the dimensions of the object.

Volume = Height * Width * Length

Volume = $3.0cm*15.0cm*5.0cm$

= $225cm^{3}$

Now we can substitute the Mass and the Volume to the density equation.

Density = [Mass/Volume]

= $\frac{24grams}{225cm^{3} }$

= $0.1067$ grams per cubic centimeter. (Rounded off to 4th decimal)

4 0

3 years ago

Round 17.62114537 to the nearest tenth

harina [27]

Answer:

17.6

Step-by-step explanation: Rounded to the nearest 0.1 or

the Tenths Place.

7 0

3 years ago

The hypotenuse of a right angled triangle is 2√13 cm . If the smaller side is increased by 2 cm and the larger side is increased

White raven [17]

Statement:

The hypotenuse of a right angled triangle is 2√13 cm. If the smaller side is increased by 2 cm and the larger side is increased by 3 cm, the new hypotenuse will be √117 cm.

To find out:

The length of the larger side of the right angled triangle.

Solution:

Let us consider x as the smaller side and y as the larger side.

Then, in the right angled triangle,

x² + y² = (2√13)² ...(I) [By Pythagoras Theorem]

Now, if the smaller side is increased by 2 cm, then the smaller side will be (x + 2).

And if the larger side is increased by 3 cm, then the larger side will be (y + 3).

Then, in the new right angled triangle,

(x + 2)² + (y + 3)² = (√117)² [By Pythagoras Theorem]

or, x² + 2 × 2 × x + 2² + y² + 2 × 3 × y + 3² = (√117)²

or, x² + 4x + 4 + y² + 6y + 9 = (√117)²

or, x² + y² + 4x + 6y + 13 = (√117)²

Now, put the value of x² + y² from equation (I),

or, (2√13)² + 4x + 6y + 13 = (√117)²

or, (2 × 2 × √13 × √13) + 4x + 6y + 13 = (√117 × √117)

or, 52 + 4x + 6y + 13 = 117

or, 4x + 6y = 117 - 52 - 13

or, 4x + 6y = 52

or, 4x = 52 - 6y

or, x = $\frac {(52 - 6y)}{4}$ ...(II)

Now, put the value of x of equation (II) in (I),

x² + y² = (2√13)²

or, $\frac {(2704-624y +36y²)}{16}$ + y² = 52

or, $\frac {(2704-624y +36y² + 16y²)}{16}$ = 52

or, 52y²-624y + 2704 = 52 × 16

or, 52y² - 624y + 2704 - 832 = 0

or, 52y² - 624y + 1872 = 0

or, 52(y² - 12y + 36) = 0

or, y²-12y +36 = 0 ÷ 52

or, y²-12y +36 = 0

or, (y)² - 2 × 6 × y + (6)² = 0

or, (y - 6)² = 0

or, y - 6=0

or, y = 6

We have taken y as the length of the larger side of the right angled triangle.

So, the length of the larger side is 6 cm.

Answer:

6 cm

Hope you could understand.

If you have any query, feel free to ask.

6 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago