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Papessa [141]
2 years ago
14

What is the simple interest on $360 borrowed for 30 months at 12.3%?

Mathematics
2 answers:
vladimir2022 [97]2 years ago
5 0
$110.70 is the answer good luck!
astra-53 [7]2 years ago
3 0

Answer:

$110.70

Step-by-step explanation:

<em>I=PRT</em>

P (principal)=360

R(rate)=12.3%

T(time)=30 months (2 and half years)

I=360*12.3%*2.5

=$110.70

Just to note 12.3% is also 0.123 we sometimes change it into a decimal.

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Do 4x and 15 + x have the same value when x is 5? How do you know?
alexandr402 [8]

Answer:

yes

Step-by-step explanation:

If x = 5,

4x=4*5=20

15+x=15+5=20

They're both the same.

4 0
2 years ago
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Convert to an improper fraction. Type in your answer with the negative in the numerator. -8 3/4 =
Sergeu [11.5K]

Answer:

-32/4

Step-by-step explanation:

-8 times 4= -32 put -32 over 4

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3 years ago
The hypotenuse of a right triangle measures 16 cm and one of its legs measures 11 cm.
Zarrin [17]

Answer:

11.6cm

Step-by-step explanation:

a^2+b^2=c^2

11^2+b^2=16^2

121+b^2=256

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6 0
2 years ago
Geometry proofs. Prove Angle B = Angle C
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Step-by-step explanation:

8 0
2 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, -90^{\circ}

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

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\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

7 0
3 years ago
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