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3 years ago

Which is equivalent to the fraction 11 over 6

Mathematics

1 answer:

den301095 [7]3 years ago

4 0

1 5/6 as an improper fraction or about 1.833 as a number.

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Try this B E Given: BELO, BOLE Prove ABEL ALOB Complete the Proof Statements 1. 2. BO=LE 3. 4. ∆BEL =∆LOB REASONS 1.GIVEN 2. 3.

AnnZ [28]

The given congruency of the sides $\mathbf{\overline{BE}}$ and $\mathbf{\overline{LO}}$ and $\mathbf{\overline{BO}}$ and $\mathbf{\overline{LE}}$ as well

as the congruency of the common side $\mathbf{\overline{BL}}$ gives.

ΔBEL ≅ ΔLOB by SSS congruency postulate

<h3>Which values correctly completes the table?</h3>

The completed two column proof is presented as follows;

Statement ${}$ Reasons

1. $\underline{\overline{BE} \cong \overline{LO}}$ ${}$ 1. Given

2. $\overline{BO}$ ≅ $\overline{LE}$ ${}$ 2.

3. $\underline{\overline{BL} \cong \overline{BL}}$ ${}$ 3. <u>Reflexive property of congruency</u>

4. ΔBEL ≅ ΔLOB ${}$ 4. <u>SSS congruency postulate</u>

Side-Side-Side, SSS, congruency postulate states that if three sides of

one triangle are congruent to three sides of another triangle, the two

triangles are congruent.

Learn more about different congruency postulates here:

brainly.com/question/1495556

7 0

2 years ago

A print shop charges a fixed amount per photocopy and gives a 10% discount off the total cost of the photocopies. The total cost

givi [52]

Answer:

the number of photocopies made

Step-by-step explanation:

The wording "a function of <em>the number of photocopies made</em>" tells you that "the number of photocopies made" is the independent variable.

___

The wording "y is a function of x" means x is the independent variable.

3 0

3 years ago

What is decrease £20 by 15%

Charra [1.4K]

$15\%\cdot\pounds20=0.15\cdot\pounds20=\pounds3\\
\pounds20-\pounds3=\pounds17$

3 0

2 years ago

Read 2 more answers

The decimal 0.26 as a fraction

Nitella [24]

Simplified: 13/50
Not Simplified: 26/100

4 0

3 years ago

Finding Derivatives Implicity In Exercise,Find dy/dx implicity.<br> x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

5 0

3 years ago