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4 years ago

Compare the values of the underline digits (2 is underline in both numbers in problem 1)

Mathematics

1 answer:

Arturiano [62]4 years ago

3 0

Answer with Step-by-step explanation:

1.We are given

2783 and 7283

We have to compare the values of the underline digits

Place value of 2 in 2783 =2000 because 2 is at thousand place

2 in 7283 is at hundred place

Therefore, the place value of 2 in 7283=200

Therefore, the value of 2 in 2783 is 10 times the value of 2 in 7283.

2.We have to compare the values of 7 in both numbers

Place of 7 in 503497=one's

Therefore, place value of 7 =7

7 in 26475 is at tens place

Therefore, the place value of 7 in 26475=70

Hence, the value of 7 in 26475 is 10 times the value of 7 in 503497.

3.We have to compare the values of 4 in both given numbers

Place of 4 in 34258=Thousand

Place value of 4 in 34258=4000

Place of 4 in 47163=Ten thousand

Place value of 4 in 47163=40000

Hence, the value of 4 in 47163 is 10 times the value of 4 in 34258.

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The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

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3 years ago

Quadrilateral GHand JK has vertices G(2, 3), H(8, 2), J(6, 8), K(3, 6). It is transformed according to the rule T(–4, –5). What

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(2,-3) because the 3 becomes a negative when transforming.

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3 years ago

Figure A is a scale image of figure B. Figure A maps to figure B with a scale factor of 0.80. What is the value of x?

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Answer:

Step-by-step explanation:

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3 years ago

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3 years ago

Write the equation in point slope intercept form of the line that passes through (6,-11) and is parallel to the graph of y=-2/3x

Alina [70]

Answer:

y+11=-2/3(x-6)

Step-by-step explanation:

To find the parallel line, we would just plug in the numbers in the point-slope form from what we're given. We have y=-2/3x+12 and (6,-11).

Point-slope form:

y-y1=m(x-x1) → y+11=-2/3(x-6)

m represents the slope, which is -2/3 in this situation.

y1 represents the y coordinate, which is -11 in this situation. However, when we plug in negative numbers in a point-slope form, we would do the opposite of the negative number, which is to make it a positive in point-slope form.

x1 represents the x coordinate, which is 6 in this situation.

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