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3 years ago

6

Simplify 6 minus the square root of 25 over 4 I would appreciate some help :)

1 answer:

marshall27 [118]3 years ago

8 0

6-5 (the square root of 25) is 1, so the answer is 1/4 or 0.25

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Solve the equation and show all work please show steps!!<br> m/9 - 17=21

masya89 [10]

Okay. The question is m/9 - 17 = 21. First off, we will add 17 to both sides. -17 + 17 cancels out. 21 + 17 is 38. That leaves m/9 = 38. Multiply each side by 9 to isolate the m, because m is the numerator and 9 is the denominator. m/9 * 9/1 cancels out. 38 * 9 is 342. Let's check this. 342/9 is 38. 38 - 17 is 21. 21 = 21. There. x = 342.

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3 years ago

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Please help<br> Regroup and write your answer in simplest form

garik1379 [7]

First you should write 2 3/4 as a decimal, 2.75 then multiply 2.75 by 2 to get 5.5. 5.5 equals 5 1/2.
Ms. Addison needs 5 1/2 cups of flour.

7 0

3 years ago

Ignore this i don’t know how to delete jt

saveliy_v [14]

Answer:

ok

Step-by-step explanation:

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3 years ago

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Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

Find the midpoint of AB for: A(7,0)|and B(0,3)

liberstina [14]

Answer:

$\boxed {\boxed {\sf (\frac {7}{2}, \frac{3}{2}) \ or \ (3,5, 1.5) }}$

Step-by-step explanation:

The midpoint is essentially a point with the average of the 2 x-coordinates and the 2 y-coordinates.

The formula is:

$(\frac {x_1+x_2}{2}, \frac{y_1+y_2}{2})$

We are given two points: A (7,0) and B (0, 3). Remember points are written as (x, y).

Therefore,

$x_1= 7 \\y_1=0 \\x_2=0 \\x_2=3$

Substitute the values into the formula.

$(\frac {7+0}{2}, \frac{0+3}{2})$

Solve the numerators first.

$(\frac {7}{2}, \frac{3}{2})$

The midpoint can be left like this because the fractions are reduced, but it can be written as decimals too.

$(3.5, 1.5)$

3 0

3 years ago