How can I make this exercise?? Please Help

Mathematics

1 answer:

Maru [420]3 years ago

8 0

x = 3 or x = 1

note that log 5 is usually taken to mean $log_{10}$ 5

using the law of logarithms

logx - logy = log ( $\frac{x}{y}$ )

the right hand side can be expressed as

log 10000 - log 16 = log ( $\frac{10000}{16}$ ) = log 625

divide both sides by log 5

x² - 4x + 7 = $\frac{log625}{log5}$ = 4

subtract 4 from both sides

x² - 4x + 3 = 0 ← in standard form

(x - 3 )(x - 1) = 0

equate each factor to zero and solve for x

x - 3 = 0 ⇒ x = 3

x - 1 = 0 ⇒ x = 1

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14. The diagonals of square ABCD intersect at E. If AE = 2, find the perimeter of ABCD.

lutik1710 [3]

Answer:

B, $8\sqrt{2}$

Step-by-step explanation:

Diagonals of a square are congruent so AC= 4

using pythagorean theorem we can then do $x^{2} +x^{2}$ = $4^{2}$

$2x^{2} = 16\\x^{2} =8\\x=\sqrt{8} \\x=2\sqrt{2}$

Then for perimeter we times the $2\sqrt{2}$ by 4 and get

$8\sqrt{2}$

Hope this helps and please mark brainliest!!

8 0

2 years ago

1 3/4 − 4/5 = 35 twentieths − what fraction???<br><br>Plss help will give brainliest :)

7nadin3 [17]

let's firstly convert the mixed fractions to improper fractions and then to do away with the denominators, let's multiply both sides by the LCD of all denominators.

$\stackrel{mixed}{1\frac{3}{4}}\implies \cfrac{1\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{7}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{4}-\cfrac{4}{5}=\cfrac{35}{20}-\boxed{?}\implies \stackrel{\textit{multipling both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{7}{4}-\cfrac{4}{5} \right)=20\left( \cfrac{35}{20}-\boxed{?} \right)} \\\\\\ 35-16=35-20\boxed{?}\implies 19=35-20\boxed{?}\implies -16=-20\boxed{?} \\\\\\ \cfrac{-16}{-20}=\boxed{?}\implies \cfrac{4}{5}=\boxed{?}$