All categories

2 years ago

Wym has a packet of 50 biscuits she gives half of them to her friend and then eats 8 how many are left

Mathematics

1 answer:

Zanzabum2 years ago

8 0

Answer:

17 biscuits

Step-by-step explanation:

Since she starts with 50 and gives away half which is 25, that will leave her with 25 biscuits herself. If she then eats 8 of them, that will leave her with 17 since 25-8=17.

You might be interested in

Please help for online school

Stels [109]

Answer:

Step-by-step explanation:

The diagram shows that the one triangle can be divided into two equal right triangles. Because of this, you can use the Pythagorean Theorem to solve this problem. a and b are the two sides of the angles, and c is the hypotenuse.

The given lengths are 5 as the hypotenuse and 3 as one length. (You have 3 as a given length because the two triangles have a length of 6 on one side. 6/2 = 3)

a² + b² = c²

a² = c² - b²

a² = 5² - 3²

a² = 25² - 9²

a² = 16

a = 4

4 0

2 years ago

3/4 (x - 3) - 5 = 1/8 ( x - 1 )

KonstantinChe [14]

Answer:

x = - $\frac{57}{5}$

Step-by-step explanation:

$\frac{3}{4}$ (x - 3) - 5 = $\frac{1}{8}$ (x - 1)

Multiply through by 8 to clear the fractions

6(x - 3) - 40 = x - 1 ← distribute parenthesis and simplify left side

6x - 18 - 40 = x - 1

6x - 58 = x - 1 ( subtract x from both sides )

5x - 58 = - 1 ( add 58 to both sides )

5x = - 57 ( divide both sides by 5 )

x = - $\frac{57}{5}$

3 0

2 years ago

How do I find the inverse of this problem?

raketka [301]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Inverse of real function.

Let's consider here, g(n) = y ,

so we get as,

$y = \sqrt[3]{ \frac{n - 1}{2} }$

no, cubing the power both side we get as,

${y}^{3} = \frac{n - 1}{2}$

now,

$2 {y}^{3} = n - 1$

so finally, we get as,

$n = 2 {y}^{3} + 1$

hence,here, n = inverse of the g(n) function.

so,

g^-1 (n) = 2y^3+1.

7 0

2 years ago

Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}

valentinak56 [21]

By definition, we have

$|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}$

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that $p+2\geq 0 \iff p \geq -2$