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3 years ago

Write ab in terms of a and b

Mathematics

2 answers:

baherus [9]3 years ago

8 0

Answer:

a multiple b

Step-by-step explanation:

a x b = ab

eg. a = 5 , b = 2

ab = 5x2

adelina 88 [10]3 years ago

8 0

ab in terms of a and b would be a x b, or a multiplied by b.

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x and y are integers between 0 and 10 that satisfy the equation shown below. x2=y⎯⎯√+3 What values for x and y satisfy the equat

8_murik_8 [283]

x2=y⎯⎯√+3
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3 years ago

9x^2 - 12x + 4 = 0 Solve for x

hram777 [196]

Answer:

x = 2/3

Step-by-step explanation:

9x2−12x+4=0

Step 1: Factor left side of equation.

(3x−2)(3x−2)=0

Step 2: Set factors equal to 0.

3x−2=0 or 3x−2=0

x= 2 /3

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3 years ago

Read 2 more answers

Find an exact solution for √5-1/x = √5/2 Then find the approximate solution to the nearest tenth.

viktelen [127]

Step-by-step explanation:

$\frac{ \sqrt{5 - 1 } }{x} = \frac{ \sqrt{5} }{2} \\ \\ x \sqrt{5 } = 2 \sqrt{5} - 2 \\ x = 2 - \frac{2 \sqrt{5} }{5}$

3 0

3 years ago

12.) Write y = 1/6 x + 7 in standard form using integers.

Trava [24]

Option A

-x + 6y = 42 is the standard form

Solution:

Given that we have to write $y = \frac{1}{6}x+7$ in standard form

The standard form of an equation is Ax + By = C

In this kind of equation, x and y are variables and A, B, and C are integers

Given equation is:

$y = \frac{1}{6}x+7$

Let us convert the above equation into standard form

$y = \frac{x}{6} + 7$

$y = \frac{x}{6} + \frac{7}{1}$

Make the denominator same in R.H.S

$y = \frac{x}{6} + \frac{7 \times 6}{1 \times 7}$

Solve the above equation

$y = \frac{x}{6} + \frac{42}{6}$

$y = \frac{x+42}{6}$

Move 6 from R.H.S to L.H.S

$6y = x + 42$

Bring all the terms to one side, leaving only constant on R.H.S

$-x + 6y = 42$

The above equation is of form Ax + By = C

Thus option A is correct

3 0

3 years ago

Paper rectangle ABCD is folded in such a way that the fold passes through point B, and so A coincides with point H where H ∈ CD.

aivan3 [116]

The answer is BC = 38.22 cm.

Step-by-step explanation:

We have, ∠BKD = 120° ,BK = 28 cm, Draw a perpendicular from point K on BC let it intersect at point M. In right angled ΔBMK, ∠BKM=30° and BK = 28 cm

sin30° = perpendicular/hypotenuse

1/2 = BM/BK

1/2 = BM/28

BM= 14 cm

Now , In right angled ΔBMK ,

cos30° = base/hypotenuse

√3/2 = MK/28

MK = 14√3 = 24.22 cm

KMCD is a square MK = MC = 24.22 cm

also, BC = BM + MC , putting values of BM & MC we get :

BC = 14 cm + 24.22 cm

BC = 38.22 cm.

5 0

3 years ago