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3 years ago

Which of the following fractions has the same value as the decimal number 0.65?

Mathematics

1 answer:

boyakko [2]3 years ago

4 0

Answer:

13/20

Step-by-step explanation:

To find this, divide 13 by 20 to get .65

Hope this helped!

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Help me please!!!!!!!!!!!

tangare [24]

Answer:

m∠B = 107°

Step-by-step explanation:

The given geometric figure is a deltoid or quadrilateral with two pairs of equal adjacent sides whose diagonals are perpendicular.

The longer diagonal divides two opposite angles into two equal parts, which means:

m∠A = 2 · 32 = 64°

m∠C = 2 · 41 = 82

Also the angles m∠B ≅ m∠D = x

The total sum of the inner angles in the quadrilateral is 360°, which means:

2 · x = 360 - ( 64 + 82) = 360 - 146 = 214 => x = 214/2 = 107°

m∠B = 107°

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3 years ago

Solve for the brainlist

Sergio039 [100]

Answer:

x = 10.0

Step-by-step explanation:

Using the geometric mean formula:

x projection

----------------- = ---------------

entire base x

Here, the projection is 5 and the entire base is (20 + 5). So:

x 5

------------ = ------------

25 x

Cross multiply:

x² = 25 (5)

x² = 100

√x² = √100

x = 10

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3 years ago

mei has eight jars of soup each jar of soup contains 300 milliliters of soup what is the smallest pot mei can use

vichka [17]

300 x 8 = 2400 ml is the smallest pot

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3 years ago

NEED HELP ASAP !!! Please

Sergio039 [100]

I do not know what is your question.

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3 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)

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3 years ago