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1 year ago

12

A certain number when divided by 4 gives a remainder 3.The same number when divided by 3 gives a remainder of two.Which of the f

ollowing can be the number?

1 answer:

klio [65]1 year ago

6 0

Answer:

2

Step-by-step explanation:

let n=4q+3

then 2n =8q+6

=4(2q+1)+2

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Triangle N P O is shown. Line O N extends through point M to form exterior angle P N M. Angle N P O is 38 degrees. Angle P O N i

Aleksandr-060686 [28]

Answer: x>38

Step-by-step explanation:

By the exterior angle theorem, x=39+38=77.

4 0

2 years ago

Write the converse of the following statement. Then determine its validity.

madam [21]

The converse and the validity of the statement is:

If a parallelogram is a rectangle, then the diagonals are congruent.
TRUE.

The statement is given as:

If the diagonals of a parallelogram are congruent, then is the parallelogram is a rectangle

Considering the following statement

If a then b

Its converse is:

If b then a

So, the converse of the given statement is:

If a parallelogram is a rectangle, then the diagonals are congruent.

The diagonals of a rectangle are true.

Hence, the validity is true.

So, the true statement is (b)

Read more about converse statements at:

brainly.com/question/4421013

5 0

2 years ago

LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

2 years ago

Help please and thank you

aksik [14]

Remember
$(a^b)^c=a^{bc}$
and
$\frac{x^a}{x^b}=x^{a-b}$

so

$\frac{(7^2)^5}{7^{-6}}=$
$\frac{7^{2*5}}{7^{-6}}=$
$\frac{7^{10}}{7^{-6}}=$
$7^{10-(-6)}=$
$7^{10+6}=$
$7^{16}$

the answer is D

5 0

3 years ago

Is the square root of 4 rational

WITCHER [35]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers