All categories

3 years ago

Please help will mark brainly

Mathematics

1 answer:

katrin2010 [14]3 years ago

3 0

Correct option d)0.61

Step-by-step explanation:

Here we have a number line from 0 to 1 . With every point at a equal distance of 0.10 from each other . Let's first mark the decimal 0.74 in number line . Now, Lets check one by one that which decimal is less than 0.74 . Remember , decimal will be less than 0.74 if point lies to the left side of 0.74.

a)0.95:

Decimal 0.95 lies to right side of 0.74 , So this decimal is not smaller than 0.74

b)0.88:

Decimal 0.88 lies to right side of 0.74 , So this decimal is not smaller than 0.74

c)0.79:

Decimal 0.79 lies to right side of 0.74 , So this decimal is not smaller than 0.74

d)0.61:

Decimal 0.61 lies to left side of 0.74 , So this decimal is smaller than 0.74

You might be interested in

478 × 965 in distributive property

kakasveta [241]

478(900) + 478(95)

Hope this helps!

4 0

3 years ago

Read 2 more answers

Which is equivalent to the complex number i^32?

Ksivusya [100]

Answer:

where i^2 = -1 (minus one)

So, i^2 = -1

i^4 = (i^2)^2 = (-1)^2 = -1 X-1 = 1

like ;

i^32 = (i^2)^16 = (-1)^16 = -1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X-1 X -1 X

= 1

Step-by-step explanation:

4 0

3 years ago

Which function represents an exponential

Leviafan [203]

I think its A!! But im not sure

6 0

3 years ago

Three girls cut out decimal models of hundredths. Amber doesn't shade any of her models. Grace qualifies three fifths of a model

Varvara68 [4.7K]

Answer:

$Amber = 0$

$Grace = 0.60$

$Marsha =1.70$

Step-by-step explanation:

Given

Decimal Grid = Hundredths

$Amber = None$

$Grace = \frac{3}{5}th$

$Marsha = 1 + \frac{7}{10}th$

To solve for each:

We have:

For Amber:

$Amber = 0 * 100$

$Amber = 0$

For Grace:

$Grace = \frac{3}{5}th$

$Grace = \frac{3}{5} * 100$

$Grace = \frac{300}{5}$

$Grace = 60$

This means that Grace shaded 60 of the hundredth models.

Decimal shaded is:

$Grace = 60 * \frac{1}{100}$

$Grace =\frac{60}{100}$

$Grace = 0.60$

For Marsha:

$Marsha = 1 + \frac{7}{10}th$

$Marsha = \frac{10 + 7}{10}th$

$Marsha = \frac{17}{10}th$

$Marsha = 1.7th$

Multiply this by 100, to get the number of shaded

$Marsha = 1.7 * 100$

$Marsha = 170$

This means that Marsha shaded 170 of the hundredth models.

Decimal shaded is:

$Marsha = \frac{170}{100}$

$Marsha =1.70$

5 0

2 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

2 years ago

Read 2 more answers