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2 years ago

9d + 10c how many terms are in this expression

Mathematics

1 answer:

bogdanovich [222]2 years ago

6 0

Answer:

I think 2

Step-by-step explanation:

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What is 3/8 equal to in decimals..?

MrRissso [65]

Answer: 0.375

Step-by-step explanation:

Hey there! If you have any questions feel free to leave them in the comments below.

To find the answer you would divide the numerator by the denominator. $\frac{3}{8}$ would be equal to 0.375

~I hope I helped you :)~

3 0

3 years ago

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A number increased by 24 is 112. what is the number

dexar [7]

I believe the correct number is 88.

8 0

2 years ago

Which statement is NOT true? A. All isosceles triangles are also equilateral triangles. B. All equilateral triangles are also is

Alecsey [184]

Hello there!

The statement that would NOT be true would be option A. All isosceles triangles are also equilateral triangles. The rest of the statements would be TRUE.

Hope this helps and have a great day! :)

5 0

3 years ago

Please Identify the congruent triangles in the figure.

Tpy6a [65]

The missing steps are each right angles and $\angle R \cong \angle O$ .

Solution:

Step 1: Given data:

$\overline {P Q} \cong \overline{M N}$

$\overline{Q R} \cong \overline{N O}$

$\overline{P R} \cong \overline{M O}$

Step 2: In the two polygons,

$\angle Q =90^\circ$ and $\angle N =90^\circ$

$\angle Q \cong \angle N$ (Each right angle)

Step 3: Given

$\angle P \cong \angle M$

Step 4: By third angle theorem,

If two angles in one triangle are congruent to the two angles in the other triangle, then the third angles in the triangles also congruent.

$\angle R \cong \angle O$

Step 5: By the definition of congruent polygons,

If two same shape polygons have all the angles are congruent and all the corresponding sides are congruent then the polygons are congruent.

Hence $\Delta P Q R \cong \Delta M N O$ .

Therefore the missing steps are each right angles and $\angle R \cong \angle O$ .

5 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$