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3 years ago

In △ABC, AB = BC = 20 and DE ≈ 9.28. Approximate BD. (made a second one since trollers ruined my last one :/ )

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

Hi! the answer is 536

marishachu [46]3 years ago

7 0

Answer:

approximately 5.36

Step-by-step explanation

20 - 9.28 = 10.72

BD = EC

10.72/2 = 5.36

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Huan's wage is £10.40 per hour l. he receives a 10% pay rise

r-ruslan [8.4K]

Answer:

Step-by-step explanation:

10% of 10.40=1.04

10.40+1.04=11.44

8 0

2 years ago

A clothing store has the sign shown in the shop window.

kicyunya [14]

Pani would pay $59 if she gets a discount of $3 on each shirt and $3 on each jeans

Step-by-step explanation:

Given

Price of shirt = $12

Price of jeans = $19

Price of shirt after $3 discount = 12-3 = $9

Price of jeans after 3 discount = 19-3 = $16

The total money Pani has to pay will be:

$9(3)+16(2)\\=27+32\\=\$59$

Pani would pay $59 if she gets a discount of $3 on each shirt and $3 on each jeans

Keywords: Discount, percent

Learn more about discount at:

#LearnwithBrainly

8 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$